(A) The sum of all probabilities in a probability distribution must be $1$.$\sum P(X=x) = 0.1 + k + 2k + 2k + k = 1$$0.1 + 6k = 1 \implies 6k = 0.9 \i

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(A) The sum of all probabilities in a probability distribution must be $1$.$\sum P(X=x) = 0.1 + k + 2k + 2k + k = 1$$0.1 + 6k = 1 \implies 6k = 0.9 \implies k = 0.15$The probability distribution is:$X$$0$$1$$2$$3$$4$$P(X)$$0.1$$0.15$$0.3$$0.3$$0.15$$1$. Probability of studying at least two hours $(X \geq 2)$:$P(X=2) + P(X=3) + P(X=4) = 0.3 + 0.3 + 0.15 = 0.75$$2$. Probability of studying exactly two hours $(X=2)$:$P(X=2) = 0.3$$3$. Probability of studying at most two hours $(X \leq 2)$:$P(X=0) + P(X=1) + P(X=2) = 0.1 + 0.15 + 0.3 = 0.55$

Class 12 Mathematics Probability Probability distribution

Medium

Let $X$ denote the number of hours you study during a randomly selected school day. The probability that $X$ can take the values $x$ has the following form,where $k$ is some unknown constant.
$P(X=x) = \begin{cases} 0.1, & \text{if } x=0 \\ kx, & \text{if } x=1 \text{ or } 2 \\ k(5-x), & \text{if } x=3 \text{ or } 4 \\ 0, & \text{otherwise} \end{cases}$
What is the probability that you study at least two hours? Exactly two hours? At most two hours?

A
$0.75, 0.3, 0.55$
B
$0.55, 0.3, 0.75$
C
$0.3, 0.55, 0.75$
D
$0.75, 0.55, 0.3$

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Probability

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Probability distribution

$A$ random variable $X$ has the probability distribution
$\begin{array}{|c|c|c|c|c|c|c|}\hline X=x_i & 1 & 2 & 3 & 4 & 5 & 6 \\\hline P(X=x_i) & 0.2 & 0.3 & 0.12 & 0.1 & 0.2 & 0.08 \\\hline \end{array}$
If $A=\{x_i \mid x_i \text{ is a prime number}\}$ and $B=\{x_i \mid x_i < 4\}$ are two events,then $P(A \cup B) = $

EasyAP EAMCET 2019

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In a game,a man wins $Rs. 100$ if he gets $5$ or $6$ on a throw of a fair die and loses $Rs. 50$ for getting any other number on the die. If he decides to throw the die either until he gets a five or a six or to a maximum of three throws,then his expected gain/loss (in rupees) is

DifficultJEE MAIN 2019

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$A$ random variable $X$ has a Poisson distribution with mean $\lambda = 2$. Then $P(X > 1.5)$ is equal to:

DifficultAIEEE 2005

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$A$ player tosses two coins. He wins $Rs. 10$ if $2$ heads appear,$Rs. 5$ if one head appears,and $Rs. 2$ if no head appears. Find the variance of the winning amount.

MediumMHT CET 2025

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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:

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

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

MediumAIEEE 2004

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

Similar Questions

In a game,a man wins $Rs. 100$ if he gets $5$ or $6$ on a throw of a fair die and loses $Rs. 50$ for getting any other number on the die. If he decides to throw the die either until he gets a five or a six or to a maximum of three throws,then his expected gain/loss (in rupees) is

$A$ random variable $X$ has a Poisson distribution with mean $\lambda = 2$. Then $P(X > 1.5)$ is equal to:

$A$ player tosses two coins. He wins $Rs. 10$ if $2$ heads appear,$Rs. 5$ if one head appears,and $Rs. 2$ if no head appears. Find the variance of the winning amount.

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

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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:

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

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