Here, three coins are tossed once so the possible outcomes or sample space (S) consist of

S=

Now,

A: ‘three heads’

A= (HHH)

B: “two heads and one tail”

B= (HHT, THH, HTH)

C: ‘three tails’

C= (TTT)

D: a head shows on the first coin

D= (HHH, HHT, HTH, HTT)

(i) Mutually exclusive

A ∩ B = (HHH) ∩ (HHT, THH, HTH)

= ϕ

Since there is no common element between A&B,

they are mutually exclusive

A ∩ C = (HHH) ∩ (TTT) = ϕ

Since here is no common element,

So, A and C are mutually exclusive.

A ∩ D = (HHH) ∩ (HHH, HHT, HTH, HTT)

= (HHH) ≠ ϕ

Since there is a common element in A and D

So they are not mutually exclusive

B ∩ C = (HHT, HTH, THH) ∩ (TTT) = ϕ

Since there is no common element in B & C, so they are mutually exclusive.

B∩ D = (HHT, THH, HTH) ∩ (HHH, HHT, HTH, HTT)

= (HHT, HTH) ≠ ϕ

Since there are common elements in B & D,

so, they not mutually exclusive.

C ∩ D = (TTT) ∩ (HHH, HHT, HTH, HTT) = ϕ

Since there is no common element in C & D,

So they are not mutually exclusive.

(ii) Simple event

An event is said to be simple if it has only one sample point in its sample space.

Here A = (HHH)

C = (TTT)

Both A & C have only one element,

so they are simple events.

(iii) Compound events

Is an event has more than one sample point of a sample space , it is called as compound event.

Here B= (HHT, HTH, THH)

D= (HHH, HHT, HTH, HTT)

Both B & D have more than one element,

So, they are compound events.