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
$\omega_{1}$	$\frac{1}{12}$
$\omega_{2}$	$\frac{1}{12}$
$\omega_{3}$	$\frac{1}{6}$
$\omega_{4}$	$\frac{1}{6}$
$\omega_{5}$	$\frac{1}{6}$
$\omega_{6}$	$\frac{3}{2}$

$X = x$	$0$	$1$	$2$	$3$
$P(X = x)$	$K$	$K + \frac{1}{7}$	$2K$	$\frac{2}{5}$

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

$A$ 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 = $

Similar Questions

If the probability function of a random variable $X$ is defined by $P(X=k) = a \left( \frac{k+1}{2^k} \right)$ for $k = 0, 1, 2, 3, 4, 5$,then the probability that $X$ takes a prime value is

For the random variable $X$ with the probability distribution given by the table:
$X = x$ $0$ $1$ $2$ $3$
$P(X = x)$ $K$ $K + \frac{1}{7}$ $2K$ $\frac{2}{5}$

The mean of $X$ is:

If a random variable $X$ has the following probability distribution,then its variance is nearly:
$X=x$ $-3$ $-2$ $-1$ $0$ $1$ $2$ $3$
$P(X=x)$ $0.05$ $0.1$ $2K$ $0$ $0.3$ $K$ $0.1$

If a discrete random variable $X$ is defined as follows:
$P(X=x) = \begin{cases} \frac{k(x+1)}{5^x}, & x=0, 1, 2, \ldots \\ 0, & \text{otherwise} \end{cases}$
then $k=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module