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(B) For a fair coin,the probability of heads is $p = 1/2$ and tails is $q = 1/2$. The random variable $X$ follows a binomial distribution $B(n, 1/2)$.

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(B) For a fair coin,the probability of heads is $p = 1/2$ and tails is $q = 1/2$. The random variable $X$ follows a binomial distribution $B(n, 1/2)$.$P(X=k) = \binom{n}{k} (1/2)^n$.Given that $P(X=4)$,$P(X=5)$,and $P(X=6)$ are in arithmetic progression,we have $2P(X=5) = P(X=4) + P(X=6)$.Substituting the binomial probabilities: $2 \binom{n}{5} (1/2)^n = \binom{n}{4} (1/2)^n + \binom{n}{6} (1/2)^n$.Dividing by $(1/2)^n$,we get $2 \binom{n}{5} = \binom{n}{4} + \binom{n}{6}$.Using the formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$,we have $2 \frac{n!}{5!(n-5)!} = \frac{n!}{4!(n-4)!} + \frac{n!}{6!(n-6)!}$.Dividing by $n!$ and multiplying by $6!(n-4)!$: $2 	imes 6(n-4) = 6 	imes 5 + (n-4)(n-5)$.$12n - 48 = 30 + n^2 - 9n + 20$.$n^2 - 21n + 98 = 0$.$(n-7)(n-14) = 0$.Thus,$n = 7$ or $n = 14$. The largest value of $n$ is $14$.

$X$ denotes 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 arithmetic progression,the largest value of $n$ is

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