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)= ^{n}C_{r} p^{r} q^{n}

P(x=4)= ^{n}C_{4} p^{4} q^{n-4} = ^{n}C_{4} p^{n}

P(x=5)= ^{n}C_{5} p^{5} q^{n-5} = ^{n}C_{5} p^{n}

P(x=6)= ^{n}C_{6} p^{6} q^{n-6} = ^{n}C_{6} p^{n}

Since they are in AP;

So,

p(x=5)=

2 p(x=5) = p(x=4)+ p(x=5)

2^{n}C_{5} p^{n} = ^{n}C_{4} p^{n} +^{n}C_{6} p^{n}

4n – 16 = 55n – 225

51n = 209

n= 4

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