(A) Since the sum of all probabilities in a probability distribution is $1$,we have:$\sum_{x=0}^7 P(X=x) = 1$$0 + k + 2k + 2k + 3k + k^2 + 2k^2 + (7k^

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(A) Since the sum of all probabilities in a probability distribution is $1$,we have:$\sum_{x=0}^7 P(X=x) = 1$$0 + k + 2k + 2k + 3k + k^2 + 2k^2 + (7k^2 + k) = 1$$10k^2 + 9k - 1 = 0$$(10k - 1)(k + 1) = 0$Since $k \geq 0$,we have $k = \frac{1}{10}$.Now,we need to find $P(X \geq 6)$:$P(X \geq 6) = P(X=6) + P(X=7)$$P(X \geq 6) = 2k^2 + (7k^2 + k) = 9k^2 + k$Substituting $k = \frac{1}{10}$:$P(X \geq 6) = 9\left(\frac{1}{10}\right)^2 + \frac{1}{10} = \frac{9}{100} + \frac{10}{100} = \frac{19}{100}$.

Class 12 Mathematics Probability Probability distribution

Easy

If a random variable $X$ has the following probability distribution values,then $P(X \geq 6)$ has the value:
$X$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$
$P(X)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2 + k$

MHT CET 2024 All MHT CET

A
$\frac{19}{100}$
B
$\frac{81}{100}$
C
$\frac{9}{100}$
D
$\frac{91}{100}$

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If a random variable $X$ has the p.d.f. $f(x) = \begin{cases} \frac{k}{x^2+1} & , \text{if } 0 < x < \infty \\ 0 & , \text{otherwise} \end{cases}$,then the c.d.f. of $X$ is:

MediumMHT CET 2025

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$A$ random variable $X$ has the following probability distribution:
$X = 1, 2, 3, 4, 5$
$P(X) = 0.1, 0.2, 0.3, 0.2, 0.2$
For the events $E = \{X \text{ is a prime number}\}$ and $F = \{X < 4\}$, find $P(E \cup F)$.

DifficultMHT CET 2026

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$A$ random variable $X$ has the following probability distribution:
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$P(X=x)$ $0.15$ $0.23$ $0.12$ $0.10$ $0.20$ $0.08$ $0.07$ $0.05$

For the events $E = \{X \text{ is a prime number}\}$ and $F = \{X < 4\}$,find $P(E \cup F)$.

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

MediumMHT CET 2024

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$A$ fair $n$ $(n > 1)$ faced die is rolled repeatedly until a number less than $n$ appears. If the mean of the number of tosses required is $\frac{n}{9}$,then $n$ is equal to

DifficultJEE MAIN 2023

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Suppose that a random variable $X$ follows a Poisson distribution. If $P(X=1) = P(X=2)$, then $P(X=5)$ is equal to:

MediumAP EAMCET 2010

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

If a random variable $X$ has the following probability distribution values,then $P(X \geq 6)$ has the value:$X$$0$$1$$2$$3$$4$$5$$6$$7$$P(X)$$0$$k$$2k$$2k$$3k$$k^2$$2k^2$$7k^2 + k$

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If a random variable $X$ has the p.d.f. $f(x) = \begin{cases} \frac{k}{x^2+1} & , \text{if } 0 < x < \infty \\ 0 & , \text{otherwise} \end{cases}$,then the c.d.f. of $X$ is:

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