(A) For a probability distribution,the sum of all probabilities must be equal to $1$.$\sum P(X) = 1$$0 + 2p + 2p + 3p + p^2 + 2p^2 + 7p^2 + 2p = 1$Com

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(A) For a probability distribution,the sum of all probabilities must be equal to $1$.$\sum P(X) = 1$$0 + 2p + 2p + 3p + p^2 + 2p^2 + 7p^2 + 2p = 1$Combining the terms,we get:$10p^2 + 9p = 1$$10p^2 + 9p - 1 = 0$Factoring the quadratic equation:$10p^2 + 10p - p - 1 = 0$$10p(p + 1) - 1(p + 1) = 0$$(10p - 1)(p + 1) = 0$This gives $p = \frac{1}{10}$ or $p = -1$.Since the probability $P(X)$ must be non-negative,$p$ must be such that all $P(X) \geq 0$. Thus,$p = -1$ is rejected.Therefore,$p = \frac{1}{10}$.

Class 12 Mathematics Probability Probability distribution

Easy

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

The value of $p$ is:

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

MHT CET 2024 All MHT CET

A
$\frac{1}{10}$
B
$\frac{1}{30}$
C
$\frac{1}{100}$
D
$\frac{3}{20}$

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Probability

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Probability distribution

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Which of the following cannot be a valid assignment of probabilities for outcomes of sample space $S = \{\omega_{1}, \omega_{2}, \omega_{3}, \omega_{4}, \omega_{5}, \omega_{6}, \omega_{7}\}$?

Outcome Probability

$\omega_{1}$ $0.1$

$\omega_{2}$ $0.2$

$\omega_{3}$ $0.3$

$\omega_{4}$ $0.4$

$\omega_{5}$ $0.5$

$\omega_{6}$ $0.6$

$\omega_{7}$ $0.7$

Medium

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If the probability function of a random variable $X$ is defined by $P(X=k) = a \left( \frac{k+1}{2^k} \right)$ for $k = 0, 1, 2, 3, 4, 5$,then the probability that $X$ takes a prime value is

DifficultAP EAMCET 2018

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If the mean of a Poisson distribution is $\frac{1}{2}$, then the ratio of $P(X=3)$ to $P(X=2)$ is

MediumAP EAMCET 2002

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The range of a random variable $X$ is $\{1, 2, 3, \ldots\}$ and $P(X=x) = \frac{c^x}{x!}$ for $x = 1, 2, 3, \ldots$. Then the value of $c$ is

MediumAP EAMCET 2021

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An executive in a company makes on an average $5$ telephone calls per hour at a cost of $Rs. 2$ per call. The probability that in any hour the cost of the calls exceeds a sum of $Rs. 4$ is

MediumTS EAMCET 2018

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Outcome	Probability
$\omega_{1}$	$0.1$
$\omega_{2}$	$0.2$
$\omega_{3}$	$0.3$
$\omega_{4}$	$0.4$
$\omega_{5}$	$0.5$
$\omega_{6}$	$0.6$
$\omega_{7}$	$0.7$

$A$ random variable $X$ has the following probability distribution:$X$$0$$1$$2$$3$$4$$5$$6$$7$$P(X)$$0$$2p$$2p$$3p$$p^2$$2p^2$$7p^2$$2p$The value of $p$ is:

Similar Questions

If the probability function of a random variable $X$ is defined by $P(X=k) = a \left( \frac{k+1}{2^k} \right)$ for $k = 0, 1, 2, 3, 4, 5$,then the probability that $X$ takes a prime value is

If the mean of a Poisson distribution is $\frac{1}{2}$, then the ratio of $P(X=3)$ to $P(X=2)$ is

The range of a random variable $X$ is $\{1, 2, 3, \ldots\}$ and $P(X=x) = \frac{c^x}{x!}$ for $x = 1, 2, 3, \ldots$. Then the value of $c$ is

An executive in a company makes on an average $5$ telephone calls per hour at a cost of $Rs. 2$ per call. The probability that in any hour the cost of the calls exceeds a sum of $Rs. 4$ is

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

The value of $p$ is: