A plane figure with ten straight sides and angles i.e.,
There are some coins in the shape of decagon, which are the real life examples of Decagon.
A plane figure with ten straight sides and angles i.e.,
There are some coins in the shape of decagon, which are the real life examples of Decagon.
Solution:
The y-coordinate of every point on the x-axis is 0.
Therefore, the equation of the x-axis is y = 0.
The x-coordinate of every point on the y-axis is 0.
Therefore, the equation of the y-axis is y = 0.
Solution:
We know that the equation of the line passing through point , whose slope m is
(y – y_{0}) = m( x – x_{0})
Thus, the equation of the line passing through point (–4, 3), whose slope is ^{1}/_{2}, is
(y – 3) = ^{1}/_{2} (x + 4)
2(y – 3) = (x + 4)
2y – 6 = (x + 4)
x – 2y + 10 = 0
Solution:
We know that the equation of the line passing through point (x_{0}, y_{0}), whose slope m is
(y – y_{0}) = m( x – x_{0})
Thus, the equation of the line passing through point (0, 0), whose slope is m , is
(y – 0) = m(x – 0)
y = mx
Solution:
The slope of the line that inclines with the x-axis at an angle of 75° is m = tan 75°
m = tan(45°+ 30°) = ^{tan45°+tan30°}/_{1 – tan45°tan30°} = [1+^{1}/_{√3}]/ [1-^{1}/_{√3}] = [√3+^{1}/_{√3}]/ [√3 – ^{1}/_{√3}] = ^{(√3+1)}/ _{(√3+1)}
We know that the equation of the line passing through point (x_{0}, y_{0}) ,whose slope is m, is
(y – y_{0}) = m( x – x_{0})
Thus, if a line passes though and inclines with the x-axis at an angle of 75°, then the equation of the line is given as
(y – 2√3) = ^{(√3+1)}/ _{(√3-1)} x (x – 2)
(y – 2√3) (√3-1) = (√3+1)(x – 2)
y(√3-1) – 2√3(√3-1) = x(√3+1) – 2(√3+1)
x(√3+1) – y(√3-1)= 2√3 + 2 – 6 + 2√3
x(√3+1) – y(√3-1)= 4√3 – 4
x(√3+1) – y(√3-1)= 4(√3 – 1)
Solution:
It is known that if a line with slope m makes x-intercept d, then the equation of the line is given as y = m(x – d)
For the line intersecting the x-axis at a distance of 3 units to the left of the origin, d = –3.
The slope of the line is given as m = –2
Thus, the required equation of the given line is y = –2 [x – (–3)] y = –2x – 6
i.e., 2x + y + 6 = 0
Solution:
It is known that if a line with slope m makes y-intercept c, then the equation of the line is given as y = mx + c
Here, c = 2 and m = tan 30° = ^{1}/_{√3}
Thus, the required equation of the given line is
y = ^{1}/_{√3}x + 2
y = ^{x+2√3}/_{√3}
√3y = x+2√3
x – √3y + 2√3 = 0
Solution:
It is known that the equation of the line passing through points (x_{1}, y_{1}) and (x_{2}, y_{2}) is
(y – y_{1})= (y_{2} – y_{1})/(x_{2} – x_{1}) x (x – x_{1})
Therefore, the equation of the line passing through the points (–1, 1) and (2, –4) is
(y – 1)= ^{-5}/_{3} x (x + 1)
3(y – 1)= -5(x + 1)
3y – 3 = -5x – 5
5x + 3y + 2 = 0
Solution:
If p is the length of the normal from the origin to a line and ω is the angle made by the normal with the positive direction of the x-axis, then the equation of the line is given by
xcos ω + y sin ω = p.
Here, p = 5 units and ω = 30°
Thus, the required equation of the given line is x
cos 30° + y sin 30° = 5
x ^{√3}/_{2} + y. ^{1}/_{2} = 5
√3x + y = 10
Solution:
It is given that the vertices of ∆PQR are P (2, 1), Q (–2, 3), and R (4, 5).
Let RL be the median through vertex R.
Accordingly, L is the mid-point of PQ.
By mid-point formula, the coordinates of point L are given by
(^{2-2}/_{2} , ^{1+3}/_{2}) = (0, 2)
It is known that the equation of the line passing through points (x_{1}, y_{1}) and (x_{2}, y_{2}) is
(y – y_{1})= (y_{2} – y_{1})/(x_{2} – x_{1}) x (x – x_{1})
Therefore, the equation of the line passing through the points (4, 5) and (0, 2) is
(y – 5)= ^{2-5}/_{0-4} x (x – 4)
(y – 5)= ^{-3}/_{-4} x (x – 4)
4(y – 5)= 3(x – 4)
4y – 20 = 3x – 12
3x – 4y + 8 = 0
Thus, the required equation of the median through vertex R is 3x – 4y + 8 = 0.
Solution:
The slope of the line joining the points (2, 5) and (–3, 6) is
m = ^{6 – 5}/_{-3-2} = ^{1}/_{-5}
We know that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other.
Therefore, slope of the line perpendicular to the line through the points (2, 5) and (–3, 6)
^{1}/_{m} = – ^{1}/(-^{1}/_{5}) = 5
Now, the equation of the line passing through point (–3, 5), whose slope is 5, is
(y – 5) = 5(x + 3)
(y – 5) = 5x + 15
5x – y + 20 = 0
Solution:
According to the section formula, the coordinates of the point that divides the line segment joining the points (1, 0) and (2, 3) in the ratio 1: n is given by
(^{n(1)+1(2)}/_{1+n} , ^{n(0)+1(3)}/_{1+n}) = (^{n+2}/_{n+1} , ^{3}/_{n+1})
The slope of the line joining the points (1, 0) and (2, 3) is
m = =^{3-0}/_{2-1} = 3
We know that two non-vertical lines are perpendicular to each other if and only if their slopes are negative reciprocals of each other.
Therefore, slope of the line that is perpendicular to the line joining the points (1, 0) and (2, 3) = – ^{1}/_{m} = – ^{1}/_{3}
Now, the equation of the line passing through (^{n+2}/_{n+1} , ^{3}/_{n+1}) and whose slope is –^{1}/_{3 }is given
(y – ^{3}/_{n+1}) = – ^{1}/_{3} (x – ^{n+2}/_{n+1})
3[(n+1)y – 9] = – (n+1)x + n + 2
(1+ n)x + 3(1 + n)y = n + 1
Solution:
The equation of a line in the intercept form is
^{x}/_{a }+ ^{y}/_{b} = 1 —————(1)
Here, a and b are the intercepts on x and y axes respectively.
It is given that the line cuts off equal intercepts on both the axes. This means that a = b.
Accordingly, equation (1) reduces to
^{x}/_{a }+ ^{y}/_{a} = 1
x + y = a ————–(2)
Since the given line passes through point (2, 3), equation (2) reduces to 2 + 3 = a
⇒ a = 5
On substituting the value of a in equation (ii), we obtain x + y = 5, which is the required equation of the line.
Solution:
The equation of a line in the intercept form is
^{x}/_{a }+ ^{y}/_{b} = 1 ——————-(1)
Here, a and b are the intercepts on x and y axes respectively.
Given that a + b = 9
⇒ b = 9 – a ————–(2)
From equations (1) and (2), we obtain
^{x}/_{a }+ ^{y}/_{9-a} = 1 ———-(3)
It is given that the line passes through point (2, 2). Therefore, equation (3) reduces to
^{2}/_{a }+ ^{2}/_{9-a} = 1
2(^{9-a+a}/_{a(9-a)}) = 1
^{18}/_{9a-a}^{2} = 1
a^{2} – 9a + 18 = 0
a^{2} – 6a – 3a + 18 = 0
a(a – 6) – 3(a – 6) = 0
(a – 6)(a – 3) = 0
If a = 6 and b = 9 – 6 = 3, then , the equation of the circle is
^{x}/_{6} +^{ y}/_{3} = 1
Thus, x + 2y – 6 = 0
If a = 3 and b = 9 – 3 = 6, then , the equation of the circle is
^{x}/_{3} +^{ y}/_{6} = 1
Thus, x + 2y – 6 = 0
Solution:
The slope of the line making an angle with the positive x-axis is m = tan(^{2π}/_{3}) = -√3
Now, the equation of the line passing through point (0, 2) and having a slope -√3 is
(y – 2) = – √3(x – 0)
(y – 2) = – √3x
√3x + y – 2 = 0
The slope of line parallel √3x + y – 2 = 0 is -√3 to line.
It is given that the line parallel to line √3x + y – 2 = 0 crosses the y-axis 2 units below the origin i.e., it passes through point (0, –2).
Hence, the equation of the line passing through point (0, –2) and having a slope -√3 is
y – (-2) = -√3(x – 0)
y + 2 = -√3x
√3x + y + 2 = 0
Solution:
The slope of the line joining the origin (0, 0) and point (–2, 9) is
Accordingly, the slope of the line perpendicular to the line joining the origin and point (– 2, 9) is
m_{2} = – ^{1}/_{m1} = – ^{1}/_{(-9/2)} = ^{2}/_{9}
Now, the equation of the line passing through point (–2, 9) and having a slope m_{2} is
(y – 9) = ^{2}/_{9} (x + 2)
9y – 81 = 2x + 4
2x – 9y + 85 = 0
Solution:
It is given that when C = 20, the value of L is 124.942, whereas when C = 110, the value of L is 125.134.
Accordingly, points (20, 124.942) and (110, 125.134) satisfy the linear relation between L and C.
Now, assuming C along the x-axis and L along the y-axis, we have two points i.e., (20, 124.942) and (110, 125.134) in the XY plane.
Therefore, the linear relation between L and C is the equation of the line passing through points (20, 124.942) and (110, 125.134).
(L – 124.942) = ^{125.134 – 124.942}/_{110-20 }(c – 20)
L – 124.942 = ^{0.192}/_{90} (c – 20)
L = ^{0.192}/_{90} (c – 20) + 124.942 which s the required linear equation.
Solution:
The relationship between selling price and demand is linear.
Assuming selling price per litre along the x-axis and demand along the y-axis, we have two points i.e., (14, 980) and (16, 1220) in the XY plane that satisfy the linear relationship between selling price and demand.
Therefore, the linear relationship between selling price per litre and demand is the equation of the line passing through points (14, 980) and (16, 1220).
y – 980 = ^{1220 – 980}/_{16-14} (x – 14)
y – 980 = ^{240}/_{2} (x – 14)
y – 980 = 120 (x – 14)
y = 120(17 – 14) + 980 = 360 + 980 = 1340
Thus the owner of the milk store could sell 1340 litres of milk weekly at Rs. 17/litre
Solution:
Let AB be the line segment between the axes and let P (a, b) be its mid-point.
Let the coordinates of A and B be (0, y) and (x, 0) respectively.
Since P (a, b) is the mid-point of AB,
(^{0+x}/_{2}, ^{y+0}/_{2}) = (a, b)
(^{x}/_{2} , ^{y}/_{2}) = (a, b)
^{x}/_{2} = a and ^{y}/_{2} = b
x = 2a and y = 2b
Thus, the respective coordinates of A and B are (0, 2b) and (2a, 0).
The equation of the line passing through points (0, 2b) and (2a, 0) is
(y – 2b) = ^{(0 – 2b)}/_{(2a – 0)} x (x – 0)
(y – 2b) = ^{(0 – 2b)}/_{(2a – 0)} x (x )
a(y – 2b) = -bx
ay – 2ab = -bx
bx + ay = 2ab
on dividing both sides by ab, we obtain
^{bx}/_{ab} + ^{ay}/_{ab} = ^{2ab}/_{ab}
^{x}/_{a} + ^{y}/_{b} = 2
Thus the equation of the line is ^{x}/_{a} + ^{y}/_{b} = 2
Solution:
Let AB be the line segment between the axes such that point R (h, k) divides AB in the ratio 1: 2.
Let the respective coordinates of A and B be (x, 0) and (0, y).
Since point R (h, k) divides AB in the ratio 1: 2, according to the section formula,
(h, k) = (^{1×0+2xx}/_{1+2} , ^{1xy+2×0}/_{1+2})
(h, k) = (^{2x}/_{3} , ^{y}/_{3})
h = ^{2x}/_{3} and k = ^{y}/_{3}
Therefore, the respective coordinates of A and B are (^{3h}/_{2}, 0) and (0, 3k)
Now, the equation of line AB passing through points (^{3h}/_{2}, 0) and (0, 3k) is
(y – 0) = ^{3k-0}/_{3h}(x – ^{3h}/_{2})
y = – ^{2k}/_{h} ( x – ^{3h}/_{2})
hy = -2kx + 3hk
2kx + hy = 3hk
Thus the required equation of line is 2kx + hy = 3hk
Solution:
In order to show that points (3, 0), (–2, –2), and (8, 2) are collinear, it suffices to show that the line passing through points (3, 0) and (–2, –2) also passes through point (8, 2).
The equation of the line passing through points (3, 0) and (–2, –2) is
(y – 0) = ^{(-2-0)}/_{(-2-3)} x (x – 3)
(y – 0) = ^{-2}/_{-5} x (x – 3)
5y = 2x – 6
2x – 5y = 6
It is observed that at x = 8 and y = 2
L.H.S = 2×8 – 5×2 = 16 – 10 = 6 = RHS
Therefore the line passing through points (3, 0) and (-2, -2) also passes through points (8, 2) hence points (3, 0), (-2, -2) and (8, 2) are collinear.
Solution:
Let ABCD be the given quadrilateral with vertices A (–4, 5), B (0, 7), C (5, –5), and D (–4, –2).
Now let us plot A, B, C, and D on the Cartesian plane and joining AB, BC, CD, and DA, the given quadrilateral can be drawn as
To find the area of quadrilateral ABCD, we draw one diagonal, say AC.
Thus, area (ABCD) = area (∆ABC) + area (∆ACD)
We know that the area of a triangle whose vertices are (x_{1}, y_{1}), (x_{2}, y_{2}), and (x_{3}, y_{3}) is
^{1}/_{2}|x_{1}(y_{2} – y_{3})+ x_{2}(y_{3} – y_{1})+ x_{3}(y_{1} – y_{2})|
Therefore, area of ∆ABC
= ^{1}/_{2}|-4(7+5)+0(-5-5)+5(5-7)|unit^{2}
= ^{1}/_{2}|-4(12)+5(-2)| unit^{2}
= ^{1}/_{2}|-48-10|unit^{2}
= ^{1}/_{2}|-58|unit^{2}
= ^{1}/_{2} x 58 unit^{2}
= 29 unit^{2}
Area of ∆ACD
= ^{1}/_{2}|-4(-5 + 2)+5(-2 – 5)+(-4)(5+5)|unit^{2}
= ^{1}/_{2}|-4(-3)+5(-7)-4(10)| unit^{2}
= ^{1}/_{2}|12 – 35 – 10|unit^{2}
= ^{1}/_{2}|-63|unit^{2}
= ^{1}/_{2} x 63 unit^{2}
= ^{63}/_{2} unit^{2}
Thus, area(ABCD) = (29 + ^{63}/_{2}) unit^{2} = ^{(58+63)}/_{2} unit^{2} = ^{121}/_{2} unit^{2}
Solution:
Let ABC be the given equilateral triangle with side 2a.
Accordingly, AB = BC = CA = 2a
Let us assume base BC lies along the y-axis such that the mid-point of BC is at the origin.
i.e., BO = OC = a, where O is the origin.
Now, it is clear that the coordinates of point C are (0, a), while the coordinates of point B are (0, –a).
It is known that the line joining a vertex of an equilateral triangle with the mid-point of its opposite side is perpendicular.
Hence, vertex A lies on the y-axis.
On applying Pythagoras theorem to ∆AOC, we obtain
(AC)^{2} = (OA)^{2} + (OC)^{2}
⇒ (2a)^{2} = (OA)^{2} + a^{2}
⇒ 4a^{2} – a^{2} = (OA)^{2}
⇒ (OA)^{2 }= 3a^{2}
⇒ OA =√3a
∴Coordinates of point A = (±√3a, 0)
Therefore, the vertices of the given equilateral triangle are
(0, a), (0, –a), and (√3a, 0)
or
(0, a), (0, –a), and (-√3a, 0)
(i) PQ is parallel to the y-axis,
(ii) PQ is parallel to the x-axis.
Solution:
The given points are P(x_{1}, y_{1}) and Q(x_{2}, y_{2})
(i) When PQ is parallel to the y-axis, x_{1} = x_{2}.
In this case, distance between P and Q = √[(x_{2} – x_{1})^{2}+(y_{2} – y_{1})^{2}]
= √(y_{2} – y_{1})^{2}
=|y_{2} – y_{1}|
(ii)When PQ is parallel to the x-axis, y_{1} = y_{2}.
In this case, distance between P and Q= √[(x_{2} – x_{1})^{2}+(y_{2} – y_{1})^{2}]
= √(x_{2} – x_{1})^{2}
=|x_{2} – x_{1}|
Solution:
Let (a, 0) be the point on the x axis that is equidistant from the points (7, 6) and (3, 4).
√[(7-a)^{2}+(6-0)^{2}] = √[(3-a)^{2} +(4 – 0)^{2}]
⇒ √[49+a^{2} -14a +36] = √[9+a^{2} – 6a + 16]
√[a^{2} – 14a + 85] = √[a^{2} – 6a + 25]
On squaring both sides, we obtain a^{2} – 14a + 85 = a^{2} – 6a + 25
⇒ –14a + 6a = 25 – 85
⇒ –8a = –60
a = ^{60}/_{8} = ^{15}/_{2}
Thus, the required point on the x-axis is (^{15}/_{2}, 0).
Solution:
The coordinates of the mid-point of the line segment joining the points P (0, –4) and B (8, 0) are (^{0+8}/_{2} , ^{-4+0}/_{2}) = (4, -2)
It is known that the slope (m) of a non-vertical line passing through the points (x_{1}, y_{1}) and (x_{2}, y_{2}) is given by m = (y_{2} – y_{1})/(x_{2} – x_{1}) , x_{2 }≠ x_{1}
Therefore, the slope of the line passing through (0, 0) and (4, –2) is
^{-2-0}/_{4-0} = ^{-2}/_{4} = ^{1}/_{2}
Hence, the required slope of the line is –^{1}/_{2}.
Solution:
Given: The vertices of the triangle are A (4, 4), B (3, 5), and C (–1, –1).
We know that the slope (m) of a non-vertical line passing through the points (x_{1}, y_{1}) and (x_{2}, y_{2}) is m = (y_{2} – y_{1})/(x_{2} – x_{1}) , x_{2 }≠ x_{1}
Slope of AB (m_{1}) = ^{5-4}/_{3-4} = -1
Slope of BC (m_{2}) = ^{-1-5}/_{-1-3} = ^{-6}/_{-4} = ^{3}/_{2}
Slope of CA (m_{3}) = ^{ 4+1}/_{4+1} = ^{5}/_{5} = 1
It is observed that m_{1}m_{3} = –1
This shows that line segments AB and CA are perpendicular to each other i.e., the given triangle is right-angled at A (4, 4).
Thus, the points (4, 4), (3, 5), and (–1, –1) are the vertices of a right-angled triangle.
Solution:
If a line makes an angle of 30° with the positive direction of the y-axis measured anticlockwise, then the angle made by the line with the positive direction of the x-axis measured anticlockwise is 90° + 30° = 120°.
Thus, the slope of the given line is tan 120° = tan (180° – 60°) = –tan 60° = – √3
Solution:
If points A (x, –1), B (2, 1), and C (4, 5) are collinear, then
Slope of AB = Slope of BC
^{1-(-1)}/_{2-x} = ^{5-1}/_{4-2}
^{1+1}/_{2-x} = ^{4}/_{2}
^{2}/_{2-x} = 2
2 = 4 – 2x
2x = 2
x = 1
Thus, the required value of x is 1.
Solution:
Let points (–2, –1), (4, 0), (3, 3), and (–3, 2) be respectively denoted by A, B, C, and D.
Slope of AB = ^{0+1}/_{4+2} = ^{1}/_{6}
Slope of AB = ^{2-3}/_{-3-3} = ^{-1}/_{-6}= ^{1}/_{6}
Slope of AB = Slope of CD
Thus AB and CD are parallel to each other
Now slope of BC = ^{3-0}/_{3-4} = ^{3}/_{-1} = -3
Slope of AD = ^{2+1}/_{-3+2} = ^{3}/_{-1} = – 3
⇒ Slope of BC = Slope of AD
⇒ BC and AD are parallel to each other.
Therefore, both pairs of opposite sides of quadrilateral ABCD are parallel. Hence, ABCD is a parallelogram.
Thus, points (–2, –1), (4, 0), (3, 3), and (–3, 2) are the vertices of a parallelogram.
Solution:
The slope of the line joining the points (3, –1) and (4, –2) is m = ^{-2-(-1)}/_{4-3} = -2 + 1 = -1
Now, the inclination (θ) of the line joining the points (3, –1) and (4, – 2) is given by tan θ = –1
⇒ θ = (90° + 45°) = 135°
Thus, the angle between the x-axis and the line joining the points (3, –1) and (4, –2) is 135°.
Solution:
Let m1 and m be the slopes of the two given lines such that m_{1} = 2m.
We know that if θ is the angle between the lines l1 and l2 with slopes m_{1} and m_{2}, then
1+ 2m^{2} = 3m
2m^{2} – 3m + 1 = 0
2m^{2}– 2m – m + 1 = 0
2m(m – 1) – 1(m – 1) = 0
(m – 1)(2m – 1) = 0
m = 1 or m = ^{1}/_{2}
If m = 1 then the slopes of the lines are 1 and 2
IF m = ^{1}/_{2} then the slopes of the lines are ^{1}/_{2} and 1
Hence the slopes of the lines are -1 and -2 or –^{1}/_{2} and -1 or 1 and 2 or ^{1}/_{2} and 1
Solution:
The slope of the line passing through (x_{1}, y_{1}) and (h, k) is (k – y_{1})/ (h – x_{1})
It is given that the slope of the line is m
[(k – y_{1})/ (h – x_{1})] = m
k – y_{1} = m(h – x_{1})
Solution:
IF the points A(h, 0) , B(a, b) and C(0, k) lie on a line then
Slope of AB = slope of BC
^{b-0}/_{a-h}^{ = k-h}/_{0-a}
^{b}/_{a-h}^{ = k-h}/_{-a}
-ab = (k – b)(a – h)
-ab = ka – kh – ab + bh
ka + bh = kh
On dividing both sides by kh, we obtain
^{ka}/_{kh} + ^{bh}/_{kh} = ^{kh}/_{kh}
^{a}/_{h} + ^{b}/_{k} = 1
Therefore, ^{a}/_{h} + ^{b}/_{k} = 1
Solution:
Since line AB passes through points A (1985, 92) and B (1995, 97), its slope is
^{97-92}/_{1995-1985} = ^{5}/_{10}= ^{1}/_{2}
Let y be the population in the year 2010. Then, according to the given graph, line AB must pass through point C (2010, y).
Therefore, Slope of AB = Slope of BC
^{1}/_{2 }= ^{y-97}/_{2010 – 1995}
^{1}/_{2 }= ^{y-97}/_{15}
^{15}/_{2} = y – 97
y – 97 = 7.5
y = 7.5 + 97 = 104.5
Thus the slope of the line AB is ^{1}/_{2} while in the year 2010, the population will be 104.5 crores.
S = {ω_{1}, ω_{2}, ω_{3}, ω_{4}, ω_{5}, ω_{6}, ω_{7}}
Assignment | ω_{1} | ω_{2} | ω_{3} | ω_{4} | ω_{5} | ω_{6} | ω_{7} |
(a) | 0.1 | 0.001 | 0.05 | 0.03 | 0.01 | 0.2 | 0.6 |
(b) | ^{1}/_{7} | ^{1}/_{7} | ^{1}/_{7} | ^{1}/_{7} | ^{1}/_{7} | ^{1}/_{7} | ^{1}/_{7} |
(c) | 0.1 | 0.2 | 0.3 | 0.4 | 0.5 | 0.6 | 0.7 |
(d) | -0.1 | 0.2 | 0.3 | 0.4 | -0.2 | 0.1 | 0.3 |
(e) | ^{1}/_{14} | ^{2}/_{14} | ^{3}/_{14} | ^{4}/_{14} | ^{5}/_{14} | ^{6}/_{14} | ^{15}/_{14} |
Solution:
ω_{1} | ω_{2} | ω_{3} | ω_{4} | ω_{5} | ω_{6} | ω_{7} |
0.1 | 0.001 | 0.05 | 0.03 | 0.01 | 0.2 | 0.6 |
Here each of the numbers p(ω_{i}), is positive and less than 1.
Sum of probabilities = p(ω_{1}) + p(ω_{2}) + p(ω_{3}) + p(ω_{4}) + p(ω_{5}) + p(ω_{6}) + p(ω_{7})
= 0.1 + 0.01 + 0.05 + 0.03 + 0.01 + 0.2 + 0.6 = 1
Therefore, the assignment is valid
(b)
ω_{1} | ω_{2} | ω_{3} | ω_{4} | ω_{5} | ω_{6} | ω_{7} |
^{1}/_{7} | ^{1}/_{7} | ^{1}/_{7} | ^{1}/_{7} | ^{1}/_{7} | ^{1}/_{7} | ^{1}/_{7} |
Here each of the numbers p(ω_{i}), is positive and less than 1.
Sum of probabilities = p(ω_{1}) + p(ω_{2}) + p(ω_{3}) + p(ω_{4}) + p(ω_{5}) + p(ω_{6}) + p(ω_{7})
= ^{1}/_{7} + ^{1}/_{7} + ^{1}/_{7} +^{1}/_{7} +^{1}/_{7} +^{1}/_{7} +^{1}/_{7} = 7 x ^{1}/_{7} = 1
Therefore, the assignment is valid
(c)
ω_{1} | ω_{2} | ω_{3} | ω_{4} | ω_{5} | ω_{6} | ω_{7} |
0.1 | 0.2 | 0.3 | 0.4 | 0.5 | 0.6 | 0.7 |
Here each of the numbers p(ω_{i}), is positive and less than 1.
Sum of probabilities = p(ω_{1}) + p(ω_{2}) + p(ω_{3}) + p(ω_{4}) + p(ω_{5}) + p(ω_{6}) + p(ω_{7})
= 0.1 + 0.2 + 0.3 + 0.4 + 0.5 + 0.6 + 0.7 = 2.8 ≠ 1
Therefore, the assignment is not valid
(d)
ω_{1} | ω_{2} | ω_{3} | ω_{4} | ω_{5} | ω_{6} | ω_{7} |
-0.1 | 0.2 | 0.3 | 0.4 | -0.2 | 0.1 | 0.3 |
Here each of the numbers p(ω_{1}) and p(ω_{5}) are negative.
Thus, the assignment is not valid
(e)
ω_{1} | ω_{2} | ω_{3} | ω_{4} | ω_{5} | ω_{6} | ω_{7} |
^{1}/_{14} | ^{2}/_{14} | ^{3}/_{14} | ^{4}/_{14} | ^{5}/_{14} | ^{6}/_{14} | ^{15}/_{14} |
Here we have p(ω_{7}) = ^{15}/_{14 } > 1
Thus the assignment is not valid
Solution:
When a coin is tossed twice, the sample space is given by S = {HH, HT, TH, TT}
Let A be an event of the occurrence of at least one tail
Then A = {HT, TH, TT}
P(A) = ^{Number of outcomes favourable to A}/_{Total number of possible outcomes}
= ^{n(A)}/_{n(S)} =^{3}/_{4}
(i) A prime number will appear
(ii) A number greater than or equal to 3 will appear
(iii) A number less than or equal to one will appear
(iv) A number more than 6 will appear
(v) A number less than 6 will appear.
Solution:
The sample space of the given experiment is given by
S = {1, 2, 3, 4, 5, 6}
(i) Let A be the event of the occurrence of a prime number.
Then, A = {2, 3, 5}
P(A) = ^{Number of outcomes favourable to A}/_{Total number of possible outcomes}
= ^{n(A)}/_{n(S)} =^{3}/_{6 } = ^{1}/_{2}
(ii) Let B be the event of the occurrence of a number greater than or equal to 3.
Then, B = {3, 4, 5, 6}
P(B) = ^{Number of outcomes favourable to A}/_{Total number of possible outcomes}
= ^{n(B)}/_{n(S)} =^{4}/_{6} = ^{2}/_{3}
(iii) Let C be the event of the occurrence of a number less than or equal to one.
Then, C = {1}
P(C) = ^{Number of outcomes favourable to A}/_{Total number of possible outcomes}
= ^{n(C)}/_{n(S)} =^{1}/_{6}
(iv) Let D be the event of the occurrence of a number greater than 6.
Then, D = Φ
P(D) = ^{Number of outcomes favourable to A}/_{Total number of possible outcomes}
= ^{n(D)}/_{n(S)} =^{0}/_{6} = 0
(v) Let E be the event of the occurrence of a number less than 6.
Then, E = {1, 2, 3, 4, 5}
P(E) = ^{Number of outcomes favourable to A}/_{Total number of possible outcomes}
= ^{n(E)}/_{n(S)} =^{5}/_{6}
(a) How many points are there in the sample space?
(b) Calculate the probability that the card is an ace of spades.
(c) Calculate the probability that the card is
(i) an ace
(ii) black card.
Solution:
(a)Given: a card is selected from a pack of 52 cards
Then the number of possible outcomes is 52 i.e., the sample space contains 52 elements.
Thus, there are 52 points in the sample space.
(b)Let A be the event in which the card drawn is an ace of spades.
Then, n(A) = 1
P(A) = ^{Number of outcomes favourable to A}/_{Total number of possible outcomes}
= ^{n(A)}/_{n(S)} =^{1}/_{52}
(c)
(i)Let E be the event in which the card drawn is an ace.
Since there are 4 aces in a pack of 52 cards, n(E) = 4
P(E) = ^{Number of outcomes favourable to A}/_{Total number of possible outcomes}
= ^{n(E)}/_{n(S)} =^{4}/_{52} = ^{1}/_{13}
(ii)Let F be the event in which the card drawn is black.
Since there are 26 black cards in a pack of 52 cards, n(F) = 26
P(F) = ^{Number of outcomes favourable to A}/_{Total number of possible outcomes}
= ^{n(F)}/_{n(S)} =^{26}/_{52} = ^{1}/_{2}
(i) 3
(ii) 12
Solution:
Given: The fair coin has 1 marked on one face and 6 on the other
We know, the die has six faces that are numbered 1, 2, 3, 4, 5, and 6, the sample space is given by
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Then, n(S) = 12
(i) Let A be the event in which the sum of numbers that turn up is 3.
Then, A = {(1, 2)}
P(A) = ^{Number of outcomes favourable to A}/_{Total number of possible outcomes}
= ^{n(A)}/_{n(S)} = ^{1}/_{12}
(ii) Let B be the event in which the sum of numbers that turn up is 12. Then B = {(6, 6)}
P(B) = ^{Number of outcomes favourable to B}/_{Total number of possible outcomes} = ^{n(B)}/_{n(S)} = ^{1}/_{12}
Solution:
There are four men and six women on the city council.
As one council member is to be selected for a committee at random, the sample space contains 10 (4 + 6) elements.
Let A be the event in which the selected council member is a woman.
Accordingly, n(A) = 6
P(A) = ^{Number of outcomes favourable to A}/_{Total number of possible outcomes}
= ^{n(A)}/_{n(S)} =^{6}/_{10} = ^{3}/_{5}
Solution:
Given: the coin is tossed four times,
We know, there can be a maximum of 4 heads or tails.
When 4 heads turns up Re1+ Re1+ Re1 + Re1 = Rs 4, is the gain.
When 3 heads and 1 tail turn up,
Re 1 + Re 1 + Re 1 – Rs 1.50 = Rs 3 – Rs 1.50 = Rs 1.50 is the gain.
When 2 heads and 2 tails turns up,
Re 1 + Re 1 – Rs 1.50 – Rs 1.50 = – Re 1, i.e., Re 1 is the loss.
When 1 head and 3 tails turn up,
Re 1 – Rs 1.50 – Rs 1.50 – Rs 1.50 = – Rs 3.50, i.e., Rs 3.50 is the loss.
When 4 tails turn up,
Rs 1.50 – Rs 1.50 – Rs 1.50 – Rs 1.50 = – Rs 6.00, i.e., Rs 6.00 is the loss.
There are 2^{4} = 16 elements in the sample space S, which is given by:
S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTTH, TTHH, HTHT, THTH, THHT, HTTT, THTT, TTHT, TTTH, TTTT}
n(S) = 16
The person wins Rs 4.00 when 4 heads turn up, i.e., when the event {HHHH} occurs.
Probability (of winning Rs 4.00) =^{1}/_{16}
The person wins Rs 1.50 when 3 heads and one tail turn up, i.e., when the event {HHHT, HHTH, HTHH, THHH} occurs.
Probability (of winning Rs 1.50) =^{4}/_{16} = ^{1}/_{4}
The person loses Re 1.00 when 2 heads and 2 tails turn up, i.e., when the event {HHTT, HTTH, TTHH, HTHT, THTH, THHT} occurs.
Probability (of losing Re 1.00) = ^{6}/_{16} = ^{3}/_{8}
The person loses Rs 3.50 when 1 head and 3 tails turn up, i.e., when the event {HTTT, THTT, TTHT, TTTH} occurs.
Probability (of losing Rs 3.50) = ^{4}/_{16} = ^{1}/_{4}
The person loses Rs 6.00 when 4 tails turn up, i.e., when the event {TTTT} occurs.
Probability (of losing Rs 6.00) = ^{1}/_{16}
(i) 3 heads
(ii) 2 heads
(iii) at least 2 heads
(iv) at most 2 heads
(v) no head
(vi) 3 tails
(vii) exactly two tails
(viii) no tail
(ix) at most two tails.
Solution:
When three coins are tossed once, the sample space is given by
S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}
Then, n(S) = 8
It is known that the probability of an event A is given by
P(A) = ^{Number of outcomes favourable to A}/_{Total number of possible outcomes}
= ^{n(A)}/_{n(S)}
(i) Let B be the event of the occurrence of 3 heads. Accordingly, B = {HHH}
P(B) = ^{n(A)}/_{n(S)} = ^{1}/_{8}
(ii) Let C be the event of the occurrence of 2 heads.
Then, C = {HHT, HTH, THH}
P(C) = ^{n(C)}/_{n(S)} = ^{3}/_{8}
(iii) Let D be the event of the occurrence of at least 2 heads.
Then, D = {HHH, HHT, HTH, THH}
P(D) = ^{n(D)}/_{n(S)} = ^{4}/_{8} = ^{1}/_{2}
(iv) Let E be the event of the occurrence of at most 2 heads.
Then, E = {HHT, HTH, THH, HTT, THT, TTH, TTT}
P(E) = ^{n(E)}/_{n(S)} = ^{7}/_{8}
(v) Let F be the event of the occurrence of no head.
Then, F = {TTT}
P(F) = ^{n(F)}/_{n(S)} = ^{1}/_{8}
(vi) Let G be the event of the occurrence of 3 tails.
Then, G = {TTT}
P(G) = ^{n(G)}/_{n(S)} = ^{1}/_{8}
(vii) Let H be the event of the occurrence of exactly 2 tails.
then, H = {HTT, THT, TTH}
P(H) = ^{n(H)}/_{n(S)} = ^{3}/_{8}
(viii) Let I be the event of the occurrence of no tail.
Thn, I = {HHH}
P(I) = ^{n(A)}/_{n(S)} = ^{1}/_{8}
(ix) Let J be the event of the occurrence of at most 2 tails.
Then, J = {HHH, HHT, HTH, THH, HTT, THT, TTH}
P(J) = = ^{n(J)}/_{n(S)} =^{7}/_{8}
Solution:
It is given that P(A) = 2/11.
Then, P(not A) = 1 – P(A)
= 1 – ^{2}/_{11}
= ^{9}/_{11}
(i) a vowel
(ii) an consonant
Solution:
There are 13 letters in the word ASSASSINATION.
∴ Hence, n(S) = 13
(i) There are 6 vowels in the given word.
∴ Probability (vowel) =^{6}/_{13}
(ii) There are 7 consonants in the given word.
∴ Probability (consonant) = ^{7}/_{13}
Solution:
Total number of ways in which one can choose six different numbers from 1 to 20
= ^{29}C_{6} = ^{20!}/_{6!(20-6)!} = ^{20!}/_{6!14!} = ^{20x19x18x17x16x15}/_{1.2.3.4.5.6} = 38760
Hence, there are 38760 combinations of 6 numbers.
Out of these combinations, one combination is already fixed by the lottery committee.
∴ Required probability of winning the prize in the game = ^{1}/_{38760}
(i) P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6
(ii) P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.8
Solution:
(i) P(A) = 0.5, P(B) = 0.7, P(A ∩ B) = 0.6
It is known that if E and F are two events such that E ⊂ F, then P(E) ≤ P(F).
However, here, P(A ∩ B) > P(A).
Hence, P(A) and P(B) are not consistently defined.
(ii) P(A) = 0.5, P(B) = 0.4, P(A ∪ B) = 0.8
It is known that if E and F are two events such that E ⊂ F, then P(E) ≤ P(F).
Here, it is seen that P(A ∪ B) > P(A) and P(A ∪ B) >P(B).
Hence, P(A) and P(B) are consistently defined.
P(A) | P(B) | P(A ∩B) | P(AUB) | |
(i) | ^{1}/_{3} | ^{1}/_{5} | ^{1}/_{15} | _______ |
(ii) | 0.35 | ________ | 0.25 | 0.6 |
(iii) | 0.5 | 0.35 | ________ | 0.7 |
Solution:
(i) Here P(A) = ^{1}/_{3} ; P(B) = ^{1}/_{5} ; P(A∩B)) = ^{1}/_{15} ;
We know that P(AUB)= P(A)+ P(B)+ P(A ∩B)
= ^{1}/_{3} + ^{1}/_{5} + ^{1}/_{15} =^{5+3-1}/_{15} = ^{7}/_{15}
(ii) Here P(A) = 0.35 ; P(B) = 0.25 ; P(AUB) = 0.6 ;
We know that P(AUB)= P(A)+ P(B) – P(A ∩B)
=0.6 = 0.35 + P(B) – 0.25
P(B) = 0.6 – 0.35 + 0.25
P(B) = 0.5
(iii) Here P(A) = 0.5, P(B) = 0.35 , P(AUB) = 0.7
We know that P(AUB)= P(A)+ P(B) – P(A ∩B)
0.7 = 0.5 + 0.35 – P(A ∩B)
P(A ∩B)= 0.5 + 0.35 – 0.7
P(A ∩B) = 0.15
Solution:
Here, P(A) = 3/5, P(B) = 1/5
For mutually exclusive events A and B,
P(A or B) = P(A) + P(B)
∴ P(A or B) = 3/5 + 1/5 = 4/5
(i) P(E or F)
(ii)P(not E and not F)
Solution:
Here P(E) = ^{1}/_{4} , P(F) = ^{1}/_{2} and P(E and F) = ^{1}/_{8}
(i) We know P(E or F) = P(E) + P(F) – P(E and F)
P(E or F) = ^{1}/_{4} + ^{1}/_{2} – ^{1}/_{4} = ^{2+4-1}/_{8} = ^{5}/_{8}
(ii) From (i) P(E or F) = P(EUF) = ^{5}/_{8}
(EUF)’=(E’ ∩ F’)
P(EUF)’ = P(E’ ∩ F’)
P(EUF)’ = 1 – P(EUF) = 1 – ^{5}/_{8} = ^{3}/_{8}
P(E’ ∩ F’) = ^{3}/_{8}
Thus P(not E and not F) = ^{3}/_{8}
Solution:
It is given that P (not E or not F) = 0.25
P(E’ U F’) = 0.25
P(E ∩F)’ = 0.25
Now P(E ∩F) = 1 – P(E ∩ F)’
P(E ∩F) = 1 – 0.25 = 0.75 ≠ 0
E ∩F ≠ 0
Therefore, E and F are not mutually exclusive
(i) P(not A)
(ii) P(not B)
(iii) P(A or B)
Solution:
It is given that P(A) = 0.42, P(B) = 0.48, P(A and B) = 0.16
(i) P(not A) = 1 – P(A) = 1 – 0.42 = 0.58
(ii) P(not B) = 1 – P(B) = 1 – 0.48 = 0.52
(iii) We know that P(A or B) = P(A) + P(B) – P(A and B)
∴ P(A or B) = 0.42 + 0.48 – 0.16 = 0.74
Solution:
Let A be the event in which the selected student studies Mathematics and B be the event in which the selected student studies Biology
Now,
P(A) = 40% = 40/100 = 2/5
P(B) = 30% = 30/100 = 3/10
P(A and B) = 10% =10/100 = 1/10
We know that P(A or B) = P(A) + P(B) – P(A and B)
P(A or B) = ^{2}/_{5} + ^{3}/_{10} – ^{1}/_{10 } = ^{6}/_{10} = 0.6
Thus, the probability that the selected student will be studying Mathematics or Biology is 0.6
What is the probability of passing both?
Solution:
Let A and B be the events of passing first and second examinations respectively.
Then P(A) = 0.8, P(B) = 0.7 and P(A or B) = 0.95
We know that P(A or B) = P(A) + P(B) – P(A and B)
∴ 0.95 = 0.8 + 0.7 – P(A and B)
⇒ P(A and B) = 0.8 + 0.7 – 0.95 = 0.55
Thus, the probability of passing both the examinations is 0.55.
Solution:
Let A and B be the events of passing English and Hindi examinations respectively.
Then, P(A and B) = 0.5, P(not A and not B) = 0.1, i.e.,
P(A) = 0.75
(AUB)’ = (A’ ∩ B’)
P(AUB)’ = P(A’ ∩ B’)
P(AUB) = 1 – P(AUB)’ = 1 – 0.1 = 0.9
We know that P(A or B) = P(A) + P(B) – P(A and B)
∴0.9 = 0.75 + P(B) – 0.5
⇒ P(B) = 0.9 – 0.75 + 0.5
⇒ P(B) = 0.65
Thus, the probability of passing the Hindi examination is 0.65.
(i) The student opted for NCC or NSS.
(ii) The student has opted neither NCC nor NSS.
(iii) The student has opted NSS but not NCC.
Solution:
Let A be the event in which the selected student has opted for NCC and B be the event in which the selected student has opted for NSS.
Total number of students = 60
Number of students who have opted for NCC = 30
∴ P(A) = ^{30}/_{60} = ^{1}/_{2}
Number of students who have opted for NSS = 32
P(B) = ^{32}/_{60} = ^{8}/_{15}
Number of students who have opted for both NCC and NSS = 24
P(A and B) = ^{24}/_{60} = ^{2}/_{5}
(i) We know that P(A or B) = P(A) + P(B) – P(A and B)
P(A or B) = ^{1}/_{2} + ^{8}/_{15} – ^{2}/_{5} = ^{15+16-12}/_{30} = ^{19}/_{30}
Thus, the probability that the selected student has opted for NCC or NSS is ^{19}/_{30}.
(ii)P(not A and not B) = P(A’ and B’)
=P(A’ ∩ B’)
= P(AUB)’
= 1 – P(AUB)
= 1 – P(A or B)
= 1 – ^{19}/_{30}
= ^{11}/_{30}
Therefore, the probability that the selected students has neither opted for NCC nor NSS is 11/30.
(iii) Number of students who have opted for NSS but not NCC
= n(B – A) = n(B) – n(A ∩ B) = 32 – 24 = 8
Thus, the probability that the selected student has opted for NSS but not for NCC = ^{8}/_{60} = ^{2}/_{15}
Are E and F mutually exclusive?
Solution:
When a die is rolled, the sample space is given by
S = {1, 2, 3, 4, 5, 6}
Then, E = {4} and F = {2, 4, 6}
It is observed that E ∩ F = {4} ≠ Φ Therefore, E and F are not mutually exclusive events.
(i) A: a number less than 7
(ii) B: a number greater than 7
(iii) C: a multiple of 3
(iv) D: a number less than 4
(v) E: an even number greater than 4
(vi) F: a number not less than 3
Also find A ∪B, A ∩B, B∪C, E ∩F, D ∩E, A – D, D – E, E ∩F’, F’
Solution:
When a die is thrown, the sample space is given by S ={1, 2, 3, 4, 5, 6}
Then,
(i) A = {1, 2, 3, 4, 5, 6}
(ii) B = Φ
(iii) C = {3, 6}
(iv). D = {1, 2, 3}
(v) E = {6}
(vi) F = {3, 4, 5, 6}
and
A ∪B ={1, 2, 3, 4, 5, 6}
A ∩B ={ }
B∪C = {3. 6}
E ∩F = {6 }
D ∩E = { }
A – C = {1, 2, 4, 5}
D – E = {1, 2, 3}
E ∩F’= { }
F’ = {1, 2}
A: the sum is greater than 8,
B: 2 occurs on either die
C: The sum is at least 7 and a multiple of 3.
Which pairs of these events are mutually exclusive?
Solution:
When a pair of dice is rolled, the sample space is given by
S = {(x, y): x, y = 1, 2, 3, 4, 5,6 }
= {(1, 1) , (1, 2) , (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Then
A = { (3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)}
B = {(1, 2), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (4, 2), (5, 2), (6, 2)}
C = {(3, 6), (4, 5), ( 5, 4), (6, 3), (6, 6)}
To find mutually exclusive pairs, we have
A∩B = { }
B∩C = { }
C∩A = {(3, 6), (4, 5), ( 5, 4), (6, 3), (6, 6)} ≠ { }
Hence, events A and B are mutually exclusive ; B and C are mutually exclusive.
(i) mutually exclusive?
(ii) simple?
(iii) compound?
Solution:
When three coins are tossed, the sample space is given by
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
Accordingly,
A = {HHH}
B = {HHT, HTH, THH}
C = {TTT}
D = {HHH, HHT, HTH, HTT}
We now observe that
A ∩ B =Φ, A ∩ C =Φ, A ∩ D = {HHH} ≠ Φ
B ∩ C =Φ, B ∩ D = {HHT, {HTH} ≠ Φ
C ∩ D = Φ
(i) Mutually exclusive events are A and B; event A and C; event B and C; and event C and D
(ii) If an event has only one sample point of a sample space, it is called a simple event. Thus, A and C are simple events.
(iii) If an event has more than one sample point of a sample space, it is called a compound event. Thus, B and D are compound events.
(i) Two events which are mutually exclusive.
(ii) Three events which are mutually exclusive and exhaustive.
(iii) Two events, which are not mutually exclusive.
(iv) Two events which are mutually exclusive but not exhaustive.
(v) Three events which are mutually exclusive but not exhaustive.
Solution:
When three coins are tossed, the sample space is given by
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
(i) Two events that are mutually exclusive can be
A: getting no heads and B: getting no tails
This is because sets A = {TTT} and B = {HHH} are disjoint.
(ii) Three events that are mutually exclusive and exhaustive can be
A: getting no heads
B: getting exactly one head
C: getting at least two heads
i.e., A= {TTT}
B= {HTT, THT, TTH}
C= {HHH, HHT, HTH, THH}
This is because A ∩ B = B ∩ C = C ∩ A = Φ and A U B U C = S
(iii) Two events that are not mutually exclusive can be
A: getting three heads B: getting at least 2 heads
i.e.,
A= {HHH}
B= {HHH, HHT, HTH, THH}
This is because A ∩ B = {HHH} ≠ Φ
(iv) Two events which are mutually exclusive but not exhaustive can be
A: getting exactly one head
B: getting exactly one tail
That is
A= {HTT, THT, TTH}
B= {HHT, HTH, THH}
It is because, A ∩ B =Φ, but A B ≠ S
(v) Three events that are mutually exclusive but not exhaustive can be
A: getting exactly three heads
B: getting one head and two tails
C: getting one tail and two heads
i.e.,
A= {HHH}
B= {HTT, THT, TTH}
C= {HHT, HTH, THH}
This is because A ∩ B = B ∩ C = C ∩ A = Φ, but A U B U C ≠ S
A: getting an even number on the first die.
B: getting an odd number on the first die.
C: getting the sum of the numbers on the dice ≤ 5.
Describe the events
(i) A’
(ii) not B
(iii) A or B
(iv) A and B
(v) A but not C
(vi) B or C
(vii) B and C
(viii)A∩B’∩C’ = A∩A∩C’ = A∩C’
Solution:
When two dice are thrown, the sample space is given by
s = {(x, y): x, y = 1, 2, 3, 4, 5, 6}
= {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Then
A = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)}
B = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
C = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
(i) A’ = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
(ii) B’ = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6)}
(iii) A or B = AUB = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)} = S
(iv) A and B = A ∩ B = { }
(v) A but not C = A – C
= {(2, 4), (2, 5), (2, 6), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
(vi) B or C = B U C = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3),(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
(vii) B and C = B ∩ C = {{(1, 1), (1, 2), (1, 3), (1, 4), (3, 1), (3, 2)}
(viii) C’ = {(1, 5), (1, 6), (2, 4), (2, 5), (2, 6), (3, 3), (3, 4), (3, 5), (3, 6), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
⸫A∩B’∩C’ = A∩A∩C’ = A∩C’
= {(2, 4), (2, 5), (2, 6), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
A: getting an even number on the first die.
B: getting an odd number on the first die.
C: getting the sum of the numbers on the dice ≤ 5
State true or false: (give reason)
(i) A and B are mutually exclusive
(ii) A and B are mutually exclusive and exhaustive
(iii) A = B’
(iv) A and C are mutually exclusive are mutually exclusive
(v) A and B’ are mutually exclusive.
(vi) A’, B’ and Care mutually exclusive and exhaustive.
Solution:
A = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
B = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
C = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}
(i) It is observed that A∩B = Φ
⸫A and B are mutually exclusive
Thus, the given statement is true
(ii) It is observed that A∩B = Φ and AUB = S
⸫ A and B are mutually exclusive and exhaustive
Thus the given statement is true.
(iii) It is observed that
B’ = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Thus the given statement is true.
(iv) It is observed that A∩C = {(2, 1), (2, 2), (2, 3), (2, 4)} ≠ Φ
⸫ A and C are not mutually exclusive
Thus, the given statement is false
(v) A∩B’ = A∩A = A
A∩B’≠ Φ
⸫ A and B are not mutually exclusive
Thus, the given statement is false
(vi) It can be observed that A’U B’U C = S
However B’ ∩ C = {(2, 1), (2, 2), (2, 3), (4, 1)} ≠ Φ
Therefore events A’, B’ and C are mutually exclusive and exhaustive.
Thus the given statement is false.
C ∩ A = Φ, but A U B U C ≠ S
Solution:
A coin has two faces: head (H) and tail (T).
It is given that, a coin is tossed three times, then the total number of possible outcomes is 2^{3}= 8
Thus, when a coin is tossed three times, the sample space is given by:
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
Solution:
When a die is thrown, the possible outcomes are 1, 2, 3, 4, 5, or 6.
When a die is thrown two times, the sample space is given by
S = {(x, y): x, y = 1, 2, 3, 4, 5, 6}
The number of elements in this sample space is 6 × 6 = 36, while the sample space is given by:
S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
Solution:
We know, when a coin is tossed once, there are two possible outcomes: head (H) and tail (T).
When a coin is tossed four times, the total number of possible outcomes is 2^{4}= 16
Thus, when a coin is tossed four times, the sample space is given by:
S = {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}
Solution:
A coin has two faces: head (H) and tail (T).
A die has six faces that are numbered from 1 to 6.
Thus, when a coin is tossed and a die is thrown, the sample space is given by:
S = {H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}
Solution:
We know A coin has two faces: head (H) and tail (T) ; a die has six faces that are numbered from 1 to 6.
Thus, when a coin is tossed and then a die is rolled only in case a head is shown on the coin, the sample space is given by: S = {H1, H2, H3, H4, H5, H6, T}
Solution:
Let us denote 2 boys and 2 girls in room X as B1, B2 and G1, G2 respectively. Let us denote 1 boy and 3 girls in room Y as B3, and G3, G4, G5 respectively.
Accordingly, the required sample space is given by
S = {XB1, XB2, XG1, XG2, YB3, YG3, YG4, YG5}
Solution:
A die has six faces that are numbered from 1 to 6, with one number on each face. Let us denote the red, white, and blue dices as R, W, and B respectively.
Accordingly, when a die is selected and then rolled, the sample space is given by
S = {R1, R2, R3, R4, R5, R6, W1, W2, W3, W4, W5, W6, B1, B2, B3, B4, B5, B6}
(i) What is the sample space if we are interested in knowing whether it is a boy or girl in the order of their births?
(ii) What is the sample space if we are interested in the number of girls in the family?
Solution:
(i) When the order of the birth of a girl or a boy is considered, the sample space is given by S = {GG, GB, BG, BB}
(ii) Since the maximum number of children in each family is 2, a family can either have 2 girls or 1 girl or no girl. Hence, the required sample space is S = {0, 1, 2}
Solution:
It is given that the box contains 1 red ball and 3 identical white balls. Let us denote the red ball with R and a white ball with W.
When two balls are drawn at random in succession without replacement, the sample space is given by S = {RW, WR, WW}
Solution:
A coin has two faces: head (H) and tail (T).
A die has six faces that are numbered from 1 to 6, with one number on each face.
Thus, in the given experiment, the sample space is given byS = {HH, HT, T1, T2, T3, T4, T5, T6}
Solution:
3 bulbs are to be selected at random from the lot. Each bulb in the lot is tested and classified as defective (D) or non-defective (N).
The sample space of this experiment is given by
S = {DDD, DDN, DND, DNN, NDD, NDN, NND, NNN}
Solution:
When a coin is tossed, the possible outcomes are head (H) and tail (T).
When a die is thrown, the possible outcomes are 1, 2, 3, 4, 5, or 6.
Thus, the sample space of this experiment is given by:
S = {T, H1, H3, H5, H21, H22, H23, H24, H25, H26, H41, H42, H43, H44, H45, H46, H61, H62, H63, H64, H65, H66}
Solution:
If 1 appears on the first drawn slip, then the possibilities that the number appears on the second drawn slip are 2, 3, or 4. Similarly, if 2 appears on the first drawn slip, then the possibilities that the number appears on the second drawn slip are 1, 3, or 4. The same holds true for the remaining numbers too.
Thus, the sample space of this experiment is given by S = {(1, 2), (1, 3), (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)}
Solution:
A die has six faces that are numbered from 1 to 6, with one number on each face. Among these numbers, 2, 4, and 6 are even numbers, while 1, 3, and 5 are odd numbers.
A coin has two faces: head (H) and tail (T).
Hence, the sample space of this experiment is given by:
S = {2H, 2T, 4H, 4T, 6H, 6T, 1HH, 1HT, 1TH, 1TT, 3HH, 3HT, 3TH, 3TT, 5HH, 5HT, 5TH, 5TT}
Solution:
The box contains 2 red balls and 3 black balls. Let us denote the 2 red balls as R1, R2 and the 3 black balls as B1, B2, and B3.
The sample space of this experiment is given by
S = {TR1, TR2, TB1, TB2, TB3, H1, H2, H3, H4, H5, H6}
Solution:
In this experiment, six may come up on the first throw, the second throw, the third throw and so on till six is obtained.
Hence, the sample space of this experiment is given by
S = {6, (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (1, 1, 6), (1, 2, 6), … , (1, 5, 6), (2, 1, 6), (2, 2, 6),… , (2, 5, 6), … ,(5, 1, 6), (5, 2, 6), …}
(i) direct method
(ii) method of contradiction
(iii)method of contrapositive
Solution:
p: “If x is a real number such that x^{3} + 4x = 0, then x is 0”.
q: x is a real number such that x^{3} + 4x = 0
r: x is 0.
(i) To show that statement p is true, we assume that q is true and then show that r is true.
Therefore, let statement q be true.
∴ x^{3} + 4x = 0 x (x^{2} + 4) = 0
⇒ x = 0 or x^{2}+ 4 = 0
Since x is real, it is 0.
Thus, statement r is true. i.e., x is 0.
Therefore, the given statement is true.
(ii) To show statement p to be true by contradiction, we assume that p is not true.
Let x be a real number such that x^{3} + 4x = 0 and let x is not 0.
Therefore, x^{3} + 4x = 0
x (x^{2} + 4) = 0
Then, x = 0 or x^{2} + 4 = 0
⇒ x = 0 or x^{2} = – 4
However, x is real.
Therefore, x = 0, which is a contradiction since we have assumed that x is not 0.
Thus, the given statement p is true.
(iii) To prove statement p to be true by contrapositive method, we assume that r is false and prove that q must be false.
Here, r is false implies that it is required to consider the negation of statement r.
This obtains the following statement.
∼r: x is not 0.
It can be seen that (x^{2} + 4) will always be positive.
x ≠ 0 implies that the product of any positive real number with x is not zero.
Let us consider the product of x with (x^{2} + 4).
∴ x (x^{2} + 4) ≠ 0
⇒ x^{3} + 4x ≠ 0
This shows that statement q is not true.
Thus, it has been proved that ∼r ⇒∼q
Therefore, the given statement p is true.
Solution:
The given statement can be written in the form of “if-then” as follows.
If a and b are real numbers such that a^{2} = b^{2}, then a = b.
Let p: a and b are real numbers such that a^{2} = b^{2}
q: a = b
The given statement has to be proved false. For this purpose, it has to be proved that if p, then ∼q.
To show this, two real numbers, a and b, with a^{2} = b^{2} are required such that a ≠ b.
Let a = 1 and b = –1 a^{2} = (1)^{2} = 1 and b^{2} = (– 1)^{2} = 1
∴ a^{2} = b^{2}
However, a ≠ b
Thus, it can be concluded that the given statement is false.
p: If x is an integer and x2 is even, then x is also even.
Solution:
p: If x is an integer and x^{2} is even, then x is also even.
Let q: x is an integer and x^{2} is even. r: x is even.
To prove that p is true by contrapositive method, we assume that r is false, and prove that q is also false.
Let x is not even.
To prove that q is false, it has to be proved that x is not an integer or x^{2} is not even.
x is not even implies that x^{2} is also not even.
Therefore, statement q is false. Thus, the given statement p is true.
(i) p: If all the angles of a triangle are equal, then the triangle is an obtuse angled triangle.
(ii) q: The equation x^{2} – 1 = 0 does not have a root lying between 0 and 2.
Solution:
(i) The given statement is of the form “if q then r”.
q: All the angles of a triangle are equal. r: The triangle is an obtuse-angled triangle.
The given statement p has to be proved false.
For this purpose, it has to be proved that if q, then ∼r.
To show this, angles of a triangle are required such that none of them is an obtuse angle.
It is known that the sum of all angles of a triangle is 180°. Therefore, if all the three angles are equal, then each of them is of measure 60°, which is not an obtuse angle.
In an equilateral triangle, the measure of all angles is equal. However, the triangle is not an obtuse-angled triangle.
Thus, it can be concluded that the given statement p is false.
(ii) The given statement is as follows.q:
The equation x^{2} – 1 = 0 does not have a root lying between 0 and 2.
This statement has to be proved false. To show this, a counter example is required.
Consider x^{2} – 1 = 0 x^{2} = 1 x = ± 1
One root of the equation x^{2} – 1 = 0, i.e. the root x = 1, lies between 0 and 2.
Thus, the given statement is false.
(i) p: Each radius of a circle is a chord of the circle.
(ii) q: The centre of a circle bisects each chord of the circle.
(iii)r: Circle is a particular case of an ellipse.
(iv) s: If x and y are integers such that x > y, then –x < –y.
(v)t: √11 is a rational number.
Solution:
(i) The given statement p is false.
According to the definition of chord, it should intersect the circle at two distinct points.
(ii) The given statement q is false.
If the chord is not the diameter of the circle, then the centre will not bisect that chord.
In other words, the centre of a circle only bisects the diameter, which is the chord of the circle.
(iii) The equation of an ellipse is, (x^{2})/(a^{2}) + (y^{2})/(b^{2}) = 1
If we put a = b = 1, then we obtain x^{2} + y^{2} = 1, which is an equation of a circle
Therefore, circle is a particular case of an ellipse.
Thus, statement r is true.
(iv) x > y
⇒ –x < –y (By a rule of inequality)
Thus, the given statement s is true.
(v) √11 is a prime number and we know that the square root of any prime number is an irrational number. Therefore, √11 is an irrational number.
Thus, the given statement t is false.