# 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.

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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