**Previous Exercise – Probability – Exercise 5.1 – Class X**

#### Probability – Exercise 5.2

- The probability that it will rain on a particular day is 0.64 what is the probability that it will not rain on that day?
- The probability of picking a non-defective item from a sample is
^{7}/_{12}. Find the probability of picking a defective item - If A is an event of a random experiment such that P(A) : P(À) = 6 : 15, then find (i) P(A) (ii)P(À)
- If A is an event of a random experiment such that P(A) =
^{3}/_{5}and P(B) =^{2}/_{7}then find P(AUB) - Two coins are tossed simultaneously. Find the probability that either both heads and both tails occur.
- When a dice is thrown, find the probability that either an odd number or a multiple of 4 occurs.
- One number card is chosen randomly from the number cards 1 to 25. Find the probability that it is divisible by 3 or 11.
- Two dice are thrown simultaneously. Find the probability that the sum of the numbers on the faces is not divisible by 4 or divisible by 5

#### Probability – Exercise 5.2 – Solutions:

**The probability that it will rain on a particular day is 0.64 what is the probability that it will not rain on that day?**

Solution:

Let A be the event of getting rain, then, À is the event of not getting rain.

It is given, P(A) = 0.64

We know, P(A) + P(À) = 1

⸫ P(À) = 1 – P(A) = 1 – 0.64 = 0.36

⸫ the probability of getting no rain is 0.36

**The probability of picking a non-defective item from a sample is**^{7}/_{12}. Find the probability of picking a defective item.

Solution:

Let A be the probability of picking a non-defective item from a sample

P(A) = ^{7}/_{12}

P(À) = 1 – P(A)

= 1 – ^{7}/_{12}

= ^{12 – 7}/_{12}

= ^{5}/_{12}

**If A is an event of a random experiment such that P(A) : P(À) = 6 : 15, then find (i) P(A) (ii)P(À)**

Solution:

(ii) Given A is an event of a random experiment such that P(A) : P(À) = 6 : 15

^{P(A)}/_{P(À)} = ^{6}/_{15}

15 x (1 – P(*À*)) = 6 x P(*À*) *[ since P(A) + P(À) = 1 **⇒** P(A) = 1 – P(À)]*

15 x P(A) = 6 x [1 – P(A)]

15P(A) = 6 – 6P(A)

15P(A) + 6P(A) = 6

21P(A) = 6

P(A) = ^{6}/_{21} = ^{2}/_{7}

(ii) P(À) = 1 – ^{2}/_{7} = ^{7-2}/_{7} = ^{5}/_{7}

**If A is an event of a random experiment such that P(A) =**^{3}/_{5}and P(B) =^{2}/_{7}then find P(AUB)

Solution:

Given, A is an event of a random experiment such that P(A) = ^{3}/_{5} and P(B) = ^{2}/_{7}

P(AUB) = P(A) + P(B)

= ^{3}/_{5} + ^{2}/_{7}

= ^{21+10}/_{35}

= ^{31}/_{35}

**Two coins are tossed simultaneously. Find the probability that either both heads or both tails occur.**

Solution:

S = {HH, HT, TH, TT} ⸫n(S) = 4

Let A is the event of getting both heads

Then A = {HH} and n(A) = 1

P(A) = ^{n(A)}/_{n(S)} = ^{1}/_{4}

Let B is the event of getting both tails

Then A = {TT} and n(B) = 1

P(B) = ^{n(B)}/_{n(S)} = ^{1}/_{4}

Now, the events A and B are mutually exclusive.

⸫P(AUB) = P(A) + P(B) = ^{1}/_{4} + ^{1}/_{4} = ^{1+1}/_{4} = ^{2}/_{4}

Therefore, the probability of getting either both heads and both tails is ^{2}/_{4}.

**When a dice is thrown, find the probability that either an odd number or a multiple of 4 occurs.**

Solution:

When a dice is thrown, a sample space , S = {1, 2, 3, 4, 5, 6}

⸫n(S) = 6

Let A be the event of getting odd number,

then A = {1, 3, 5}

n(A) = 3

P(A) = ^{n(A)}/_{n(S)} = ^{3}/_{6}

Let B be the event of getting a multiple of 4

then B= {4}

n(B) = 1

P(B) = ^{n(B)}/_{n(S)} = ^{1}/_{6}

Now the events A and B are mutually exclusive.

P(AUB) = P(A) + P(B)

= ^{3}/_{6} + ^{1}/_{6}

= ^{4}/_{6}

⸫the probability that either an odd number or a multiple of 4 is ^{4}/_{6}

**One number card is chosen randomly from the number cards 1 to 25. Find the probability that it is divisible by 3 or 11.**

Solution:

Number card from 1 to 25 = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25}

n(S) = 25

Let A be the event of getting the card number which is divisible by 3

A = {3, 6, 9, 12, 15, 18, 21, 25}

n(A) = 8

P(A) = ^{n(A)}/_{n(S)}= ^{8}/_{25}

Let B be the event of getting the card number which is divisible by 11

B = {11, 22}

n(B) = 2

P(B) = ^{n(B)}/_{n(S)}= ^{2}/_{25}

Now the event A and B are mutually exclusive.

⸫P(AUB) = P(A) + P(B)

= ^{8}/_{25} + ^{2}/_{25}

= ^{10}/_{25}

**Two dice are thrown simultaneously. Find the probability that the sum of the numbers on the faces is not divisible by 4 or divisible by 5**

Solution:

When two dice are thrown simultaneously, then sample space 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)}

n(S) = 36

Let A be the event of getting sum of the number on the faces divisible by 4

A = {(4, 4)}

n(A) = 1

P(A) = ^{n(A)}/_{n(S)} = ^{1}/_{36}

Let B be the event of getting sum of the number on the faces divisible by 5

B = {(5, 5) }

n(B) = 1

P(B) = ^{n(B)}/_{n(S)} = ^{1}/_{36}

Now, A and B occur mutually exclusive. Then,

P(AUB) = P(A) + P(B)

=^{1}/_{36} + ^{1}/_{36}

= ^{2}/_{36}

Therefore, the probability that the sum of the numbers on the faces is divisible by 4 or divisible by 5 is ^{2}/_{36}

Now, let C be the probability that the sum of the numbers on the faces is not divisible by 4 or divisible by 5

Thus, P(AUB) + P(C) = 1

P(C) = 1 – P(AUB)

= 1 – ^{2}/_{36}

= ^{34}/_{36}

**Next exercise – Probability – Exercise 5.3**