(B) For a probability distribution,the sum of all probabilities must be equal to $1$.Therefore,$\sum P(X) = 1$.$K + \frac{1}{6} + \frac{3}{8} + 2K + \

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(B) For a probability distribution,the sum of all probabilities must be equal to $1$.Therefore,$\sum P(X) = 1$.$K + \frac{1}{6} + \frac{3}{8} + 2K + \frac{1}{12} = 1$Combine the terms with $K$ and the constant fractions:$3K + (\frac{1}{6} + \frac{3}{8} + \frac{1}{12}) = 1$Find a common denominator for the fractions,which is $24$:$3K + (\frac{4}{24} + \frac{9}{24} + \frac{2}{24}) = 1$$3K + \frac{15}{24} = 1$$3K + \frac{5}{8} = 1$$3K = 1 - \frac{5}{8}$$3K = \frac{3}{8}$$K = \frac{1}{8}$

Class 12 Mathematics Probability Probability distribution

Easy

The random variable $X$ has the following probability distribution:
| $X$ | $8$ | $12$ | $16$ | $20$ | $24$ |
|---|---|---|---|---|---|
| $P(X)$ | $K$ | $\frac{1}{6}$ | $\frac{3}{8}$ | $2K$ | $\frac{1}{12}$ |
Then the value of $K$ is:

MHT CET 2022 All MHT CET

A
$\frac{1}{4}$
B
$\frac{1}{8}$
C
$\frac{3}{2}$
D
$\frac{1}{2}$

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Probability distribution

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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:

MediumAP EAMCET 2021

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$A$ person throws an unbiased die. If the number shown is even,he gains an amount equal to the number shown. If the number is odd,he loses an amount equal to the number shown. Then his expectation is ₹.

EasyMHT CET 2024

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State which of the following is not the probability distribution of a random variable. Give reasons for your answer.

$Y$ $-1$ $0$ $1$

$P(Y)$ $0.6$ $0.1$ $0.2$

Easy

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$A$ fair die is tossed twice in succession. If $X$ denotes the number of fours in $2$ tosses,then the probability distribution of $X$ is given by

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The p.m.f. of a random variable $X$ is given by $P(X=x) = \begin{cases} \frac{\binom{5}{x}}{2^5}, & x = 0, 1, 2, 3, 4, 5 \\ 0, & \text{otherwise} \end{cases}$. Then which of the following is not correct?

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

The random variable $X$ has the following probability distribution:| $X$ | $8$ | $12$ | $16$ | $20$ | $24$ ||---|---|---|---|---|---|| $P(X)$ | $K$ | $\frac{1}{6}$ | $\frac{3}{8}$ | $2K$ | $\frac{1}{12}$ |Then the value of $K$ is:

Similar Questions

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:

$A$ person throws an unbiased die. If the number shown is even,he gains an amount equal to the number shown. If the number is odd,he loses an amount equal to the number shown. Then his expectation is ₹.

State which of the following is not the probability distribution of a random variable. Give reasons for your answer. $Y$ $-1$ $0$ $1$ $P(Y)$ $0.6$ $0.1$ $0.2$

$A$ fair die is tossed twice in succession. If $X$ denotes the number of fours in $2$ tosses,then the probability distribution of $X$ is given by

The p.m.f. of a random variable $X$ is given by $P(X=x) = \begin{cases} \frac{\binom{5}{x}}{2^5}, & x = 0, 1, 2, 3, 4, 5 \\ 0, & \text{otherwise} \end{cases}$. Then which of the following is not correct?

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The random variable $X$ has the following probability distribution:
| $X$ | $8$ | $12$ | $16$ | $20$ | $24$ |
|---|---|---|---|---|---|
| $P(X)$ | $K$ | $\frac{1}{6}$ | $\frac{3}{8}$ | $2K$ | $\frac{1}{12}$ |
Then the value of $K$ 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:

State which of the following is not the probability distribution of a random variable. Give reasons for your answer.

$Y$ $-1$ $0$ $1$

$P(Y)$ $0.6$ $0.1$ $0.2$