A biased coin with probability p, 0

Question

(B) Let $X$ be the random variable representing the number of tosses required to get the first head. This follows a geometric distribution with parame

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(B) Let $X$ be the random variable representing the number of tosses required to get the first head. This follows a geometric distribution with parameter $p$. The probability mass function is given by $P(X = r) = (1 - p)^{r - 1} p$ for $r = 1, 2, 3, \dots$. We are interested in the event $E$ where the number of tosses is even,i.e.,$X \in \{2, 4, 6, \dots\}$. The probability $P(E)$ is the sum of the probabilities of these mutually exclusive events: $P(E) = P(X = 2) + P(X = 4) + P(X = 6) + \dots$ $P(E) = (1 - p)p + (1 - p)^3 p + (1 - p)^5 p + \dots$ This is an infinite geometric series with the first term $a = p(1 - p)$ and common ratio $r = (1 - p)^2$. The sum of an infinite geometric series is $S = \frac{a}{1 - r}$. $P(E) = \frac{p(1 - p)}{1 - (1 - p)^2} = \frac{p(1 - p)}{1 - (1 - 2p + p^2)} = \frac{p(1 - p)}{2p - p^2} = \frac{p(1 - p)}{p(2 - p)} = \frac{1 - p}{2 - p}$. Given that $P(E) = \frac{2}{5}$,we set up the equation: $\frac{1 - p}{2 - p} = \frac{2}{5}$ $5(1 - p) = 2(2 - p)$ $5 - 5p = 4 - 2p$ $5 - 4 = 5p - 2p$ $1 = 3p$ $p = \frac{1}{3}$.

Outcome	Probability
$\omega_{1}$	$0.1$
$\omega_{2}$	$0.01$
$\omega_{3}$	$0.05$
$\omega_{4}$	$0.03$
$\omega_{5}$	$0.01$
$\omega_{6}$	$0.2$
$\omega_{7}$	$0.6$

$X = x$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X = x)$	$0.15$	$0.23$	$0.10$	$0.12$	$0.20$	$0.08$	$0.07$	$0.05$

$X=x$	$-3$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X=x)$	$0.05$	$0.1$	$2K$	$0$	$0.3$	$K$	$0.1$

$A$ biased coin with probability $p, 0 < p < 1,$ of heads 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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