Answer :

since 3 coins are tossed, the sample space will consist of following possibilities

S= (HHH, HHT, HTH, HTT, THH,THT, TTH, TTT)

(i) Two events which are mutually exclusive.

Let A be the event of getting only head

A= (HHH)

And let B be the event of getting only Tail

B=(TTT)

So A ∩ B = ϕ

Since there is no common element in A& B so these two are mutually exclusive.

(ii) Three events which are mutually exclusive and exhaustive

Now,

Let A be the event of getting exactly two tails

A= (HTT, TTH,THT)

Let B be the event of getting at least two heads

B = (HHT, HTH, THH, HHH)

Let C be the event of getting only one tail

C= (TTT)

A ∩ B = (HTT, TTH,THT) ∩ (HHT, HTH, THH, HHH)

= ϕ

Since there is no common element in A and B hence they are mutually exclusive

B ∩ C = (HHT, HTH, THH, HHH) ∩ (TTT)

= ϕ

Since there is no common element in B and C

So they are mutually exclusive.

A ∩ C =( HTT, TTH,THT) ∩ (TTT) = ϕ

Since there is no common element in A and C,

So they mutually exclusive

Now, since A & B, B& C and A& C are mutually exclusive

∴ A, B, and C are mutually exclusive.

Also,

A ⋃ B ⋃ C = (HTT, TTH,THT, HHT, HTH, THH, HHH, TTT) = S

Hence A, B and C are exhaustive events.

(iii) Two events , which are not mutually exclusive

Let a be the event of getting at least two heads

A = ( HHH, HHT, THH, HTH)

Let B be the event of getting only head

B= (HHH)

Now A ∩ B = (HHH) ≠ ϕ

Since there is a common element in A & B,

So they are not mutually exclusive.

(iv) Two events which are mutually exclusive but not exhaustive

Let A be the event of getting only Head

A= (HHH)

Let b be the event of getting only tail

B = (TTT)

A ∩ B = ϕ

Since there is no common element in A & B,

These are mutually exclusive events.

But

A ⋃ B = (HHH) ∪ (TTT)

= ≠ S

Since A ∪ B ≠ S these are not exhaustive events.

(v) Three events which are mutually exclusive but not exhaustive

Let A be the event of getting only head

A = (HHH)

Let B be the event of getting only tail

B = (TTT)

Let C be the event of getting exactly two heads

C= (HHT, THH, HTH)

Now,

A ∩ B = (HHH) ∩ (TTT) = ϕ

A ∩ C = (HHH) ∩ (HHT, THH, HTH) = ϕ

B ∩ C = (TTT) ∩ (HHT, THH, HTH) = ϕ

⇒ A ∩ B = ϕ, A ∩ C = ϕ and B ∩ C = ϕ

i.e; they are mutually exclusive

also

A ∪ B ∪ C = (HHH TTT, HHT, THH, HTH) ≠ S

Since A ∪ B ∪ C ≠ S,

So, A, B and C are not exhaustive.

It`s is hence proved that A, B and C are mutually exclusive but not exhaustive.

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