A biased coin with probability p (0

Question

(B) Let $q = 1-p$ be the probability of getting a tail. The event that the first head appears on an even toss means the sequence of outcomes is $(T, H

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(B) Let $q = 1-p$ be the probability of getting a tail. The event that the first head appears on an even toss means the sequence of outcomes is $(T, H), (T, T, T, H), (T, T, T, T, T, H), \dots$ The probability of this occurring is $P = qp + q^3p + q^5p + \dots$ This is an infinite geometric series with first term $a = qp$ and common ratio $r = q^2$. Since $0 The sum of the series is $P = \frac{a}{1-r} = \frac{qp}{1-q^2}$. Given $P = \frac{2}{5}$, we have $\frac{qp}{1-q^2} = \frac{2}{5}$. Substituting $q = 1-p$, we get $\frac{(1-p)p}{1-(1-p)^2} = \frac{2}{5}$. Simplifying the denominator: $1 - (1 - 2p + p^2) = 2p - p^2 = p(2-p)$. So, $\frac{p(1-p)}{p(2-p)} = \frac{2}{5}$. Canceling $p$ (since $p 
eq 0$), we get $\frac{1-p}{2-p} = \frac{2}{5}$. Cross-multiplying: $5(1-p) = 2(2-p) \Rightarrow 5 - 5p = 4 - 2p$. $1 = 3p \Rightarrow p = \frac{1}{3}$.

$X$	$4k$	$\frac{30}{7}k$	$\frac{32}{7}k$	$\frac{34}{7}k$	$\frac{36}{7}k$	$\frac{38}{7}k$	$\frac{40}{7}k$	$6k$
$P(X)$	$\frac{2}{15}$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{1}{5}$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{1}{5}$	$\frac{1}{15}$

$A$ biased coin with probability $p$ $(0 < p < 1)$ of getting a head is tossed until a head appears for the first time. If the probability that the number of tosses required is even is $\frac{2}{5}$, then $p=$

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