**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.

**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}

**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.

**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.

**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

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

**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)}

**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

** **