Probability – Exercise 16.2 – Class XI

1. A die is rolled. Let E be the event “die shows 4” and F be the event “die shows even number”.

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.

1. A die is thrown. Describe the following 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}

1. An experiment involves rolling a pair of dice and recording the numbers that come up. Describe the following events:

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.

1. Three coins are tossed once. Let A denote the event ‘three heads show”, B denote the event “two heads and one tail show”. C denote the event “three tails show” and D denote the event ‘a head shows on the first coin”. Which events are

(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.

1. Three coins are tossed. Describe

(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

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

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

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

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

1. Two dice are thrown. The events A, B and C are as follows:

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)}

1. Two dice are thrown. The events A, B and C are as follows:

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