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Mark the correct alternative in the following:
Let X denote the number of times heads occur in n tosses of a fair coin. If P (X = 4), P(X = 5) and P(X = 6) are in AP; the value of n is
Given:
P(H)=1/2=p
q=1/2
P(x=r)= nCr pr qn
P(x=4)= nC4 p4 qn-4 = nC4 pn
P(x=5)= nC5 p5 qn-5 = nC5 pn
P(x=6)= nC6 p6 qn-6 = nC6 pn
Since they are in AP;
So,
p(x=5)=
2 p(x=5) = p(x=4)+ p(x=5)
2nC5 pn = nC4 pn +nC6 pn
4n – 16 = 55n – 225
51n = 209
n= 4