(C) We know that the sum of all probabilities in a distribution must be equal to $1$.$\sum P(X = x) = 0 + k + 2k + 2k + 3k + k^2 + 2k^2 + 7k^2 + k = 1

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(C) We know that the sum of all probabilities in a distribution must be equal to $1$.$\sum P(X = x) = 0 + k + 2k + 2k + 3k + k^2 + 2k^2 + 7k^2 + k = 1$$10k^2 + 9k - 1 = 0$$(10k - 1)(k + 1) = 0$Since $k$ must be positive for probabilities to be valid, we have $k = \frac{1}{10}$.We need to find $P(0 $P(0 Substituting $k = \frac{1}{10}$, we get $P(0 < X < 4) = 5 \times \frac{1}{10} = \frac{1}{2}$.

Class 12 Mathematics Probability Probability distribution

Medium

The probability distribution of a random variable $X$ is given below:
$X = x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$
$P(X = x)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2 + k$

Then, $P(0 < X < 4)$ is equal to:

TS EAMCET 2021 All TS EAMCET

A
$\frac{4}{5}$
B
$\frac{3}{5}$
C
$\frac{1}{2}$
D
$\frac{1}{4}$

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Probability distribution

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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 $\omega_1$ $\omega_2$ $\omega_3$ $\omega_4$ $\omega_5$ $\omega_6$
$(a)$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{6}$

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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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Let $F(x)$ be the cumulative distribution function (c.d.f.) of a continuous random variable $X$. If $F(b) = 0.7$ and $P(X > a) = 0.4$, then the value of $P(a < X < b)$ is ...

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The probability mass function of a random variable $X$ is given by $P[X = r] = \begin{cases} \frac{^n C_r}{32}, & r = 0, 1, 2, \dots, n \\ 0, & \text{otherwise} \end{cases}$. Then,$P[X \leq 2] = $

MediumMHT CET 2023

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The probability distribution of a random variable $X$ is given below:
$x$ $1$ $2$ $3$ $4$ $5$ $6$
$P(X=x)$ $a$ $a$ $a$ $b$ $b$ $0.3$

If the mean of $X$ is $4.2$, then $a$ and $b$ are respectively equal to:

MediumTS EAMCET 2017

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

Outcome	$\omega_1$	$\omega_2$	$\omega_3$	$\omega_4$	$\omega_5$	$\omega_6$
$(a)$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$

The probability distribution of a random variable $X$ is given below:$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, $P(0 < X < 4)$ is equal to:

Similar Questions

$A$ 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 ₹.

Let $F(x)$ be the cumulative distribution function (c.d.f.) of a continuous random variable $X$. If $F(b) = 0.7$ and $P(X > a) = 0.4$, then the value of $P(a < X < b)$ is ...

The probability mass function of a random variable $X$ is given by $P[X = r] = \begin{cases} \frac{^n C_r}{32}, & r = 0, 1, 2, \dots, n \\ 0, & \text{otherwise} \end{cases}$. Then,$P[X \leq 2] = $

The probability distribution of a random variable $X$ is given below:$x$$1$$2$$3$$4$$5$$6$$P(X=x)$$a$$a$$a$$b$$b$$0.3$If the mean of $X$ is $4.2$, then $a$ and $b$ are respectively equal to:

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The probability distribution of a random variable $X$ is given below:
$X = x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$
$P(X = x)$ $0$ $k$ $2k$ $2k$ $3k$ $k^2$ $2k^2$ $7k^2 + k$

Then, $P(0 < X < 4)$ is equal to:

The probability distribution of a random variable $X$ is given below:
$x$ $1$ $2$ $3$ $4$ $5$ $6$
$P(X=x)$ $a$ $a$ $a$ $b$ $b$ $0.3$

If the mean of $X$ is $4.2$, then $a$ and $b$ are respectively equal to: