(D) Since the sum of probabilities is $1$:$\frac{1}{6} + k + \frac{1}{4} + k + \frac{1}{6} = 1$$2k + \frac{7}{12} = 1 \implies 2k = \frac{5}{12} \impl

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(D) Since the sum of probabilities is $1$:$\frac{1}{6} + k + \frac{1}{4} + k + \frac{1}{6} = 1$$2k + \frac{7}{12} = 1 \implies 2k = \frac{5}{12} \implies k = \frac{5}{24}$Calculation for Variance:$\begin{array}{|c|c|c|c|} \hline x_i & P(X=x_i) & x_i P(x_i) & x_i^2 P(x_i) \\ \hline -2 & 1/6 & -1/3 & 2/3 \\ \hline -1 & 5/24 & -5/24 & 5/24 \\ \hline 0 & 1/4 & 0 & 0 \\ \hline 1 & 5/24 & 5/24 & 5/24 \\ \hline 2 & 1/6 & 1/3 & 2/3 \\ \hline \text{Total} & 1 & 0 & 21/12 \\ \hline \end{array}$$\text{Variance} = E(X^2) - [E(X)]^2$$= \frac{21}{12} - (0)^2 = \frac{7}{4}$

Class 12 Mathematics Probability Probability distribution

Easy

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

$X=x_i$ $-2$ $-1$ $0$ $1$ $2$

$P(X=x_i)$ $1/6$ $k$ $1/4$ $k$ $1/6$

The variance of this random variable is

AP EAMCET 2018 All AP EAMCET

A
$0$
B
$\frac{5}{24}$
C
$\frac{3}{24}$
D
$\frac{7}{4}$

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Probability

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The distribution of a random variable $X$ is given below:
$X=x$ $1$ $2$ $3$ $4$
$P(X=x)$ $\frac{2}{20}$ $\frac{4}{20}$ $\frac{6}{20}$ $\frac{8}{20}$

Then,the standard deviation of $X$ is

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

$X$ $1$ $2$ $3$ $4$ $5$

$P(X)$ $K$ $2K$ $2K$ $3K$ $K$

Let $p=P(1 < X < 4 \mid X < 3)$. If $5p = \lambda K$,then $\lambda$ is equal to .... .

MediumJEE MAIN 2021

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In a game,$3$ coins are tossed. $A$ person is paid ₹ $100$,if he gets all heads or all tails; and he is supposed to pay ₹ $40$,if he gets one head or two heads. The amount he can expect to win/lose on an average per game in (₹) is

MediumMHT CET 2024

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$A$ random variable $X$ has the following probability distribution:
$X = x$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$
$P(X = x)$ $0.15$ $0.23$ $0.10$ $0.12$ $0.20$ $0.08$ $0.07$ $0.05$

For the event $E = \{ X \text{ is a prime number} \}$,$F = \{ X < 4 \}$,then $P(E \cup F)$ is

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

MediumMHT CET 2024

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$A$ random variable $X$ has the following probability distribution:
$X$ $0$ $1$ $2$
$P(X)$ $\frac{25}{36}$ $k$ $\frac{1}{36}$

If the mean of the random variable $X$ is $\frac{1}{3}$,then the variance is:

$X$	$0$	$1$	$2$
$P(X)$	$\frac{25}{36}$	$k$	$\frac{1}{36}$

MediumKCET 2024

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$X=x$	$1$	$2$	$3$	$4$
$P(X=x)$	$\frac{2}{20}$	$\frac{4}{20}$	$\frac{6}{20}$	$\frac{8}{20}$

$A$ random variable $X$ has the following probability distribution: $X=x_i$ $-2$ $-1$ $0$ $1$ $2$ $P(X=x_i)$ $1/6$ $k$ $1/4$ $k$ $1/6$ The variance of this random variable is

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The distribution of a random variable $X$ is given below: $X=x$$1$$2$$3$$4$$P(X=x)$$\frac{2}{20}$$\frac{4}{20}$$\frac{6}{20}$$\frac{8}{20}$ Then,the standard deviation of $X$ is

The probability distribution of random variable $X$ is given by: $X$ $1$ $2$ $3$ $4$ $5$ $P(X)$ $K$ $2K$ $2K$ $3K$ $K$ Let $p=P(1 < X < 4 \mid X < 3)$. If $5p = \lambda K$,then $\lambda$ is equal to .... .

In a game,$3$ coins are tossed. $A$ person is paid ₹ $100$,if he gets all heads or all tails; and he is supposed to pay ₹ $40$,if he gets one head or two heads. The amount he can expect to win/lose on an average per game in (₹) is

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$A$ random variable $X$ has the following probability distribution:$X$$0$$1$$2$$P(X)$$\frac{25}{36}$$k$$\frac{1}{36}$If the mean of the random variable $X$ is $\frac{1}{3}$,then the variance is:

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

$X=x_i$ $-2$ $-1$ $0$ $1$ $2$

$P(X=x_i)$ $1/6$ $k$ $1/4$ $k$ $1/6$

The variance of this random variable is

The distribution of a random variable $X$ is given below:
$X=x$ $1$ $2$ $3$ $4$
$P(X=x)$ $\frac{2}{20}$ $\frac{4}{20}$ $\frac{6}{20}$ $\frac{8}{20}$

Then,the standard deviation of $X$ is

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

$X$ $1$ $2$ $3$ $4$ $5$

$P(X)$ $K$ $2K$ $2K$ $3K$ $K$

Let $p=P(1 < X < 4 \mid X < 3)$. If $5p = \lambda K$,then $\lambda$ is equal to .... .

$A$ random variable $X$ has the following probability distribution:
$X = x$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$
$P(X = x)$ $0.15$ $0.23$ $0.10$ $0.12$ $0.20$ $0.08$ $0.07$ $0.05$

For the event $E = \{ X \text{ is a prime number} \}$,$F = \{ X < 4 \}$,then $P(E \cup F)$ is

$A$ random variable $X$ has the following probability distribution:
$X$ $0$ $1$ $2$
$P(X)$ $\frac{25}{36}$ $k$ $\frac{1}{36}$

If the mean of the random variable $X$ is $\frac{1}{3}$,then the variance is: