A random variable X has the following probabi | vedclass

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

Question

(B) The sum of all probabilities in a probability distribution must be equal to $1$.$\sum P(X) = 1 \Rightarrow K^2 + 2K + K + 2K + 5K^2 = 1$$6K^2 + 5K

Vedclass · Accepted Answer

$\frac{23}{36}$

Vedclass · Answer

(B) The sum of all probabilities in a probability distribution must be equal to $1$.$\sum P(X) = 1 \Rightarrow K^2 + 2K + K + 2K + 5K^2 = 1$$6K^2 + 5K - 1 = 0$Factoring the quadratic equation: $(6K - 1)(K + 1) = 0$This gives $K = \frac{1}{6}$ or $K = -1$.Since the probability $P(X)$ cannot be negative,we reject $K = -1$. Thus,$K = \frac{1}{6}$.We need to find $P(X > 2) = P(X = 3) + P(X = 4) + P(X = 5)$.$P(X > 2) = K + 2K + 5K^2 = 3K + 5K^2$.Substituting $K = \frac{1}{6}$:$P(X > 2) = 3(\frac{1}{6}) + 5(\frac{1}{6})^2 = \frac{1}{2} + \frac{5}{36} = \frac{18}{36} + \frac{5}{36} = \frac{23}{36}$.

$X$	$1, 2, 3, 4, 5$
$P(X)$	$K^2, 2K, K, 2K, 5K^2$

$X = x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(x)$	$k$	$2k$	$3k$	$4k$	$4k$	$3k$	$2k$	$k$	$k$

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

$A$ random variable $X$ has the following probability distribution:
$X$ $1, 2, 3, 4, 5$
$P(X)$ $K^2, 2K, K, 2K, 5K^2$

Then $P(X > 2)$ is equal to:

Similar Questions

$A$ random variable $X$ has the following probability distribution. Find the value of $k$ and the value of $P(3 < X \leq 6)$.
$X = x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$
$P(x)$ $k$ $2k$ $3k$ $4k$ $4k$ $3k$ $2k$ $k$ $k$

The probability distribution of a random variable $X$ is given by

$X=x_i$ $0$ $1$ $2$ $3$ $4$

$P(X=x_i)$ $0.4$ $0.3$ $0.1$ $0.1$ $0.1$

Then the variance of $X$ is

$A$ 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 $P(X > 6)$.

The expected value of the sum of the two numbers obtained on the uppermost faces,when two fair dice are rolled,is

The cumulative distribution function of a continuous random variable $X$ is given by $F(x) = \frac{\sqrt{x}}{2}$ for $0 \leq x \leq 4$. Then $P[X > 1]$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module

$A$ random variable $X$ has the following probability distribution:$X$$1, 2, 3, 4, 5$$P(X)$$K^2, 2K, K, 2K, 5K^2$Then $P(X > 2)$ is equal to: