(B) The expected value $E(X)$ of a random variable $X$ is given by the formula $E(X) = \sum p_i x_i$.Using the provided table:$E(X) = (0 \times 0.35)

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(B) The expected value $E(X)$ of a random variable $X$ is given by the formula $E(X) = \sum p_i x_i$.Using the provided table:$E(X) = (0 \times 0.35) + (500 \times 0.25) + (1000 \times 0.15) + (1500 \times 0.10) + (2000 \times 0.08) + (2500 \times 0.05) + (3000 \times 0.02)$Calculating each term:$E(X) = 0 + 125 + 150 + 150 + 160 + 125 + 60$Summing the values:$E(X) = 770$Thus,the expected value of the maintenance cost is Rs. $770$.

Class 12 Mathematics Probability Probability distribution

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Rajesh has just bought a $VCR$ from Maharashtra Electronics and the shop offers an after-sales service contract for Rs. $1000$ for the next five years. Considering the experience of $VCR$ users,the following distribution of maintenance expenses for the next five years is formed:
Expenses $0$ $500$ $1000$ $1500$ $2000$ $2500$ $3000$
Probability $0.35$ $0.25$ $0.15$ $0.10$ $0.08$ $0.05$ $0.02$

The expected value of the maintenance cost is:

Expenses	$0$	$500$	$1000$	$1500$	$2000$	$2500$	$3000$
Probability	$0.35$	$0.25$	$0.15$	$0.10$	$0.08$	$0.05$	$0.02$

MHT CET 2021 All MHT CET

A
Rs. $800$
B
Rs. $770$
C
Rs. $700$
D
Rs. $900$

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$A$ random variable $X$ has the following probability distribution:
$X$ $1, 2, 3, 4, 5$
$P(X)$ $K^2, 2K, K, 2K, 5K^2$

Then $P(X > 2)$ is equal to:

DifficultMHT CET 2024

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Three defective oranges are accidentally mixed with seven good ones and on looking at them,it is not possible to differentiate between them. Two oranges are drawn at random from the lot. If $x$ denotes the number of defective oranges,then the variance of $x$ is:

DifficultJEE MAIN 2025

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$A$ player tosses two coins. He wins $Rs. 1$ if $1$ head appears,$Rs. 2$ if $2$ heads appear. But he loses $Rs. 3$ if no head appears. The mean of the prize money is

DifficultAP EAMCET 2017

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Find the probability distribution of the number of tails in the simultaneous tosses of three coins.

Medium

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If $X$ is a Poisson variate such that $P(X=1) = 2P(X=2)$,then $P(X=3)$ is equal to:

DifficultTS EAMCET 2013

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$A$ random variable $X$ has the following probability distribution:$X$$1, 2, 3, 4, 5$$P(X)$$K^2, 2K, K, 2K, 5K^2$Then $P(X > 2)$ is equal to:

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$A$ random variable $X$ has the following probability distribution:
$X$ $1, 2, 3, 4, 5$
$P(X)$ $K^2, 2K, K, 2K, 5K^2$

Then $P(X > 2)$ is equal to: