Let a random variable X take values {0, 1, 2, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The sum of probabilities must be $1$: $P(X=0) + P(X=1) + P(X=2) + P(X=3) = 1$ $p + p + q + q = 1 \implies 2p + 2q = 1 \implies p + q = \frac{1}{2}

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) The sum of probabilities must be $1$: $P(X=0) + P(X=1) + P(X=2) + P(X=3) = 1$ $p + p + q + q = 1 \implies 2p + 2q = 1 \implies p + q = \frac{1}{2}$ Now,calculate the expected value $E(X)$: $E(X) = \sum x_i P(X=x_i) = 0(p) + 1(p) + 2(q) + 3(q) = p + 5q$ Calculate the expected value $E(X^2)$: $E(X^2) = \sum x_i^2 P(X=x_i) = 0^2(p) + 1^2(p) + 2^2(q) + 3^2(q) = p + 13q$ Given $E(X^2) = 2E(X)$: $p + 13q = 2(p + 5q)$ $p + 13q = 2p + 10q$ $p = 3q$ Substitute $p = 3q$ into $p + q = \frac{1}{2}$: $3q + q = \frac{1}{2} \implies 4q = \frac{1}{2} \implies q = \frac{1}{8}$ Then $p = 3(\frac{1}{8}) = \frac{3}{8}$ Finally,calculate $8p - 1$: $8(\frac{3}{8}) - 1 = 3 - 1 = 2$

$X$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X)$	$0$	$k$	$2k$	$3k$	$3k^2$	$k^2$	$2k^2$	$7k^2+k$

Outcome	$\omega_1$	$\omega_2$	$\omega_3$	$\omega_4$	$\omega_5$	$\omega_6$
$(a)$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$

Let a random variable $X$ take values $\{0, 1, 2, 3\}$ with $P(X=0) = P(X=1) = p$,$P(X=2) = P(X=3) = q$,and $E(X^2) = 2E(X)$. Then the value of $8p - 1$ is:

Similar Questions

$A$ random variable $X$ has the following probability distribution:

$X$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$

$P(X)$ $0$ $k$ $2k$ $3k$ $3k^2$ $k^2$ $2k^2$ $7k^2+k$

Determine $P(X > 6)$.

$A$ player tosses $2$ fair coins. He wins $Rs. 5$ if $2$ heads appear,$Rs. 2$ if $1$ head appears,and $Rs. 1$ if no head appears. Then,the variance of his winning amount is

If the probability distribution is given by $P(x) = C \binom{4}{x}$ for $x = 0, 1, 2, 3, 4$,then find the value of $C$.

$A$ random variable $X$ takes values $0, 1, 2, 3, \ldots$ with probability $P(X=x) = K(x+1)\left(\frac{1}{5}\right)^x$,where $K$ is a constant. Then $P(X=0)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module