(D) For a probability distribution,the sum of all probabilities must be equal to $1$.Therefore,$k + 0 + 2k + 5k + k + 3k = 1$.$12k = 1 \Rightarrow k =

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(D) For a probability distribution,the sum of all probabilities must be equal to $1$.Therefore,$k + 0 + 2k + 5k + k + 3k = 1$.$12k = 1 \Rightarrow k = \frac{1}{12}$.We need to find $P(X \geq 4)$,which is $P(X = 4) + P(X = 5) + P(X = 6)$.$P(X \geq 4) = 5k + k + 3k = 9k$.Substituting the value of $k$,we get $P(X \geq 4) = 9 \times \frac{1}{12} = \frac{9}{12} = \frac{3}{4}$.

Class 12 Mathematics Probability Probability distribution

Easy

Given below is the probability distribution of a discrete random variable $X$:
$X = x$ $1$ $2$ $3$ $4$ $5$ $6$
$P(X = x)$ $k$ $0$ $2k$ $5k$ $k$ $3k$

Then $P(X \geq 4) = $

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

MHT CET 2020 All MHT CET

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

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For a random variable $X$,if $P(X=k) = \frac{(k+1)a}{3^k}$ for $k=0, 1, 2, \ldots$,then $a = $

MediumAP EAMCET 2017

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$A$ 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 = $

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The probability distribution of a random variable $X$ is given by the following table:
$X = x$ $1$ $2$ $3$ $\dots$ $n$
$P(X = x)$ $\frac{1}{n}$ $\frac{1}{n}$ $\frac{1}{n}$ $\dots$ $\frac{1}{n}$

Then $\operatorname{Var}(X) = $

$X = x$	$1$	$2$	$3$	$\dots$	$n$
$P(X = x)$	$\frac{1}{n}$	$\frac{1}{n}$	$\frac{1}{n}$	$\dots$	$\frac{1}{n}$

MediumMHT CET 2021

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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:

$X = x_i$	$0$	$1$	$2$	$3$	$4$
$P(X = x_i)$	$0.4$	$0.3$	$0.1$	$0.1$	$0.1$

EasyMHT CET 2024

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$A$ random variable $X$ has the following probability distribution:
$X = x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$
$P(X = x)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2 + k$

Then $F(4) = $

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

EasyMHT CET 2021

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Given below is the probability distribution of a discrete random variable $X$:$X = x$$1$$2$$3$$4$$5$$6$$P(X = x)$$k$$0$$2k$$5k$$k$$3k$Then $P(X \geq 4) = $

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For a random variable $X$,if $P(X=k) = \frac{(k+1)a}{3^k}$ for $k=0, 1, 2, \ldots$,then $a = $

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Given below is the probability distribution of a discrete random variable $X$:
$X = x$ $1$ $2$ $3$ $4$ $5$ $6$
$P(X = x)$ $k$ $0$ $2k$ $5k$ $k$ $3k$

Then $P(X \geq 4) = $

The probability distribution of a random variable $X$ is given by the following table:
$X = x$ $1$ $2$ $3$ $\dots$ $n$
$P(X = x)$ $\frac{1}{n}$ $\frac{1}{n}$ $\frac{1}{n}$ $\dots$ $\frac{1}{n}$

Then $\operatorname{Var}(X) = $

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

Then $F(4) = $