(A) Given that $F(x)$ is the c.d.f. of $X$, we have $F(x) = P(X \le x)$.We are given $F(b) = P(X \le b) = 0.7$.We are given $P(X > a) = 0.4$. Since $P

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(A) Given that $F(x)$ is the c.d.f. of $X$, we have $F(x) = P(X \le x)$.We are given $F(b) = P(X \le b) = 0.7$.We are given $P(X > a) = 0.4$. Since $P(X \le a) + P(X > a) = 1$, we have $P(X \le a) = 1 - 0.4 = 0.6$.We need to find $P(a Since $X$ is a continuous random variable, $P(a Substituting the values, $P(a < X < b) = 0.7 - 0.6 = 0.1$.

Class 12 Mathematics Probability Probability distribution

Difficult

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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A
$0.1$
B
$0.2$
C
$0.3$
D
$0.5$

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Probability

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Probability distribution

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If on an average $4$ customers visit a shop in an hour, then the probability that more than $2$ customers visit the shop in a specific hour is

DifficultTS EAMCET 2024

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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 $\omega_{1}$ $\omega_{2}$ $\omega_{3}$ $\omega_{4}$ $\omega_{5}$ $\omega_{6}$ $\omega_{7}$
Probability $-0.1$ $0.2$ $0.3$ $0.4$ $-0.2$ $0.1$ $0.3$

Easy

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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 $\omega_1$ $\omega_2$ $\omega_3$ $\omega_4$ $\omega_5$ $\omega_6$

$(b)$ $1$ $0$ $0$ $0$ $0$ $0$

Outcome	$\omega_1$	$\omega_2$	$\omega_3$	$\omega_4$	$\omega_5$	$\omega_6$
$(b)$	$1$	$0$	$0$	$0$	$0$	$0$

Easy

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For a Poisson variate $X$,if $P(X=2)=3 P(X=3)$,then the mean of $X$ is:

DifficultTS EAMCET 2003

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The cumulative distribution function of a discrete random variable $X$ is given by the following table:
$X = x$ $-4$ $-2$ $0$ $2$ $4$ $6$ $8$ $10$
$F(X = x)$ $0.1$ $0.3$ $0.5$ $0.65$ $0.75$ $0.85$ $0.90$ $1$

Then,calculate $\frac{P(X \leqslant 0)}{P(X > 0)}$.

$X = x$	$-4$	$-2$	$0$	$2$	$4$	$6$	$8$	$10$
$F(X = x)$	$0.1$	$0.3$	$0.5$	$0.65$	$0.75$	$0.85$	$0.90$	$1$

MediumMHT CET 2025

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

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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If on an average $4$ customers visit a shop in an hour, then the probability that more than $2$ customers visit the shop in a specific hour is

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 $\omega_1$ $\omega_2$ $\omega_3$ $\omega_4$ $\omega_5$ $\omega_6$ $(b)$ $1$ $0$ $0$ $0$ $0$ $0$

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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 $\omega_1$ $\omega_2$ $\omega_3$ $\omega_4$ $\omega_5$ $\omega_6$

$(b)$ $1$ $0$ $0$ $0$ $0$ $0$

The cumulative distribution function of a discrete random variable $X$ is given by the following table:
$X = x$ $-4$ $-2$ $0$ $2$ $4$ $6$ $8$ $10$
$F(X = x)$ $0.1$ $0.3$ $0.5$ $0.65$ $0.75$ $0.85$ $0.90$ $1$

Then,calculate $\frac{P(X \leqslant 0)}{P(X > 0)}$.