(B) The event $E$ is defined as $X$ being a prime number. The prime numbers in the given set are $\{2, 3, 5, 7\}$.$P(E) = P(2) + P(3) + P(5) + P(7) =

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(B) The event $E$ is defined as $X$ being a prime number. The prime numbers in the given set are $\{2, 3, 5, 7\}$.$P(E) = P(2) + P(3) + P(5) + P(7) = 0.23 + 0.12 + 0.20 + 0.07 = 0.62$.The event $F$ is defined as $X $P(F) = P(1) + P(2) + P(3) = 0.15 + 0.23 + 0.12 = 0.50$.The intersection event $E \cap F$ consists of values that are both prime and less than $4$,which are $\{2, 3\}$.$P(E \cap F) = P(2) + P(3) = 0.23 + 0.12 = 0.35$.Using the addition rule for probability,$P(E \cup F) = P(E) + P(F) - P(E \cap F)$.$P(E \cup F) = 0.62 + 0.50 - 0.35 = 0.77$.

Class 12 Mathematics Probability Probability distribution

Medium

$A$ random variable $X$ has the probability distribution as shown below. For the events $E = \{ X \text{ is a prime number} \}$ and $F = \{ X < 4 \}$,the probability $P(E \cup F)$ is:

$X$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$

$P(X)$ $0.15$ $0.23$ $0.12$ $0.10$ $0.20$ $0.08$ $0.07$ $0.05$

AIEEE 2004 All AIEEE

A
$0.5$
B
$0.77$
C
$0.35$
D
$0.87$

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Class 12

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Probability

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Probability distribution

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AIEEE 2004 Paper All AIEEE years →

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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The following is the probability distribution of $X$:

$X$ $0$ $1$ $2$ $3$

$P(X=x)$ $\frac{1+p}{5}$ $\frac{2-2p}{5}$ $\frac{2-p}{5}$ $\frac{2p}{5}$

For a minimum value of $p$,the value of $5 E(X)$ is:

$X$	$0$	$1$	$2$	$3$
$P(X=x)$	$\frac{1+p}{5}$	$\frac{2-2p}{5}$	$\frac{2-p}{5}$	$\frac{2p}{5}$

MediumMHT CET 2025

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Let $X$ be a random variable such that the probability function of a distribution is given by $P(X=0) = \frac{1}{2}$ and $P(X=j) = \frac{1}{3^j}$ for $j = 1, 2, 3, \ldots, \infty$. Then the mean of the distribution and $P(X \text{ is positive and even})$ respectively are:

MediumJEE MAIN 2021

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Face masks are supplied to a junior college in packets of $100$. If there is a chance that $1$ in $500$ face masks is defective, then the number of packets containing no defective face masks in a consignment of $10,000$ packets is:

DifficultTS EAMCET 2021

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The cumulative distribution function (c.d.f.) $F(x)$ of a discrete random variable $X$ is given by the following table:
$X$ $-3$ $-1$ $0$ $1$ $3$ $5$ $7$ $9$
$F(X=x)$ $0.1$ $0.3$ $0.5$ $0.65$ $0.75$ $0.85$ $0.90$ $1$

Then,find the value of $\frac{P[X=-3]}{P[X < 0]}$.

$X$	$-3$	$-1$	$0$	$1$	$3$	$5$	$7$	$9$
$F(X=x)$	$0.1$	$0.3$	$0.5$	$0.65$	$0.75$	$0.85$	$0.90$	$1$

EasyMHT CET 2025

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

$A$ random variable $X$ has the probability distribution as shown below. For the events $E = \{ X \text{ is a prime number} \}$ and $F = \{ X < 4 \}$,the probability $P(E \cup F)$ is: $X$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $P(X)$ $0.15$ $0.23$ $0.12$ $0.10$ $0.20$ $0.08$ $0.07$ $0.05$

Similar Questions

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:

The following is the probability distribution of $X$: $X$ $0$ $1$ $2$ $3$ $P(X=x)$ $\frac{1+p}{5}$ $\frac{2-2p}{5}$ $\frac{2-p}{5}$ $\frac{2p}{5}$ For a minimum value of $p$,the value of $5 E(X)$ is:

Let $X$ be a random variable such that the probability function of a distribution is given by $P(X=0) = \frac{1}{2}$ and $P(X=j) = \frac{1}{3^j}$ for $j = 1, 2, 3, \ldots, \infty$. Then the mean of the distribution and $P(X \text{ is positive and even})$ respectively are:

Face masks are supplied to a junior college in packets of $100$. If there is a chance that $1$ in $500$ face masks is defective, then the number of packets containing no defective face masks in a consignment of $10,000$ packets is:

The cumulative distribution function (c.d.f.) $F(x)$ of a discrete random variable $X$ is given by the following table:$X$$-3$$-1$$0$$1$$3$$5$$7$$9$$F(X=x)$$0.1$$0.3$$0.5$$0.65$$0.75$$0.85$$0.90$$1$Then,find the value of $\frac{P[X=-3]}{P[X < 0]}$.

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$A$ random variable $X$ has the probability distribution as shown below. For the events $E = \{ X \text{ is a prime number} \}$ and $F = \{ X < 4 \}$,the probability $P(E \cup F)$ is:

$X$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$

$P(X)$ $0.15$ $0.23$ $0.12$ $0.10$ $0.20$ $0.08$ $0.07$ $0.05$

The following is the probability distribution of $X$:

$X$ $0$ $1$ $2$ $3$

$P(X=x)$ $\frac{1+p}{5}$ $\frac{2-2p}{5}$ $\frac{2-p}{5}$ $\frac{2p}{5}$

For a minimum value of $p$,the value of $5 E(X)$ is:

The cumulative distribution function (c.d.f.) $F(x)$ of a discrete random variable $X$ is given by the following table:
$X$ $-3$ $-1$ $0$ $1$ $3$ $5$ $7$ $9$
$F(X=x)$ $0.1$ $0.3$ $0.5$ $0.65$ $0.75$ $0.85$ $0.90$ $1$

Then,find the value of $\frac{P[X=-3]}{P[X < 0]}$.