Q. 54.0( 33 Votes )

Three coins

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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