(B) For a probability distribution,the sum of all probabilities must be $1$: $\sum P(X=x_i) = 1$ $0 + k + 2k + 2k + 3k + k^2 + 2k^2 + (7k^2 + k) = 1$

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(B) For a probability distribution,the sum of all probabilities must be $1$: $\sum P(X=x_i) = 1$ $0 + k + 2k + 2k + 3k + k^2 + 2k^2 + (7k^2 + k) = 1$ $10k^2 + 9k - 1 = 0$ $(10k - 1)(k + 1) = 0$ Since $P(X=x) \ge 0$,$k$ must be positive,so $k = \frac{1}{10}$. Now,we need to find $P(0 $P(0 Substituting $k = \frac{1}{10}$: $P(0 < X < 6) = 8\left(\frac{1}{10}\right) + \left(\frac{1}{10}\right)^2 = \frac{8}{10} + \frac{1}{100} = \frac{80+1}{100} = \frac{81}{100} = \left(\frac{9}{10}\right)^2$.

Class 12 Mathematics Probability Probability distribution

Medium

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

$x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$

$P(X=x)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2+k$

Find the value of $P(0 < X < 6)$.

AP EAMCET 2018 All AP EAMCET

A
$\frac{9}{10}$
B
$\left(\frac{9}{10}\right)^2$
C
$\frac{3}{10}$
D
$\frac{1}{10}$

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Probability

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Probability distribution

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$A$ random variable $X$ has the following probability distribution:
$X = x_i$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $9$
$P(X = x_i)$ $10k$ $9k$ $8k$ $8k$ $6k$ $5k$ $4k$ $3k$ $k$

where $k$ is a real number. If $A = \{ x_i : x_i \text{ is a prime number} \}$ and $B = \{ x_i : x_i > 5 \}$ are two events, then $P(A \cup B) = $

MediumTS EAMCET 2022

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Let a sample space be $S = \{\omega_{1}, \omega_{2}, \ldots, \omega_{6}\}$. Which of the following assignments of probabilities to each outcome is valid?
Outcome Probability
$\omega_{1}$ $\frac{1}{12}$
$\omega_{2}$ $\frac{1}{12}$
$\omega_{3}$ $\frac{1}{6}$
$\omega_{4}$ $\frac{1}{6}$
$\omega_{5}$ $\frac{1}{6}$
$\omega_{6}$ $\frac{3}{2}$

Easy

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$A$ fair die is tossed repeatedly until a six is obtained. Let $X$ denote the number of tosses required and let $a=P(X=3)$,$b=P(X \geq 3)$ and $c=P(X \geq 6 \mid X>3)$. Then $\frac{b+c}{a}$ is equal to

DifficultJEE MAIN 2024

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$A$ random variable $X$ has the following probability distribution:
$X$ $0$ $1$ $2$ $3$ $4$ $5$ $6$
$P(X)$ $k$ $3k$ $5k$ $7k$ $9k$ $11k$ $13k$

Then find $P(X \ge 2)$.

$X$	$0$	$1$	$2$	$3$	$4$	$5$	$6$
$P(X)$	$k$	$3k$	$5k$	$7k$	$9k$	$11k$	$13k$

MediumMHT CET 2022

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If the following function is a probability density function of a random variable $X$, $f(x) = kx^2(1 - x)$ for $0 < x < 1$ and $f(x) = 0$ otherwise, then the value of $k$ is:

DifficultMHT CET 2026

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

$X = x_i$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$	$9$
$P(X = x_i)$	$10k$	$9k$	$8k$	$8k$	$6k$	$5k$	$4k$	$3k$	$k$

Outcome	Probability
$\omega_{1}$	$\frac{1}{12}$
$\omega_{2}$	$\frac{1}{12}$
$\omega_{3}$	$\frac{1}{6}$
$\omega_{4}$	$\frac{1}{6}$
$\omega_{5}$	$\frac{1}{6}$
$\omega_{6}$	$\frac{3}{2}$

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

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$A$ fair die is tossed repeatedly until a six is obtained. Let $X$ denote the number of tosses required and let $a=P(X=3)$,$b=P(X \geq 3)$ and $c=P(X \geq 6 \mid X>3)$. Then $\frac{b+c}{a}$ is equal to

$A$ random variable $X$ has the following probability distribution:$X$$0$$1$$2$$3$$4$$5$$6$$P(X)$$k$$3k$$5k$$7k$$9k$$11k$$13k$Then find $P(X \ge 2)$.

If the following function is a probability density function of a random variable $X$, $f(x) = kx^2(1 - x)$ for $0 < x < 1$ and $f(x) = 0$ otherwise, then the value of $k$ is:

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

$x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$

$P(X=x)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2+k$

Find the value of $P(0 < X < 6)$.

$A$ random variable $X$ has the following probability distribution:
$X$ $0$ $1$ $2$ $3$ $4$ $5$ $6$
$P(X)$ $k$ $3k$ $5k$ $7k$ $9k$ $11k$ $13k$

Then find $P(X \ge 2)$.