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

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