JEE MAIN Probability Distribution Previous Year Questions (PYQs) – Page 1 of 3
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Showing 26 questions
Let X be a random variable such that the probability function of a distribution is given by $P(X = 0) = {1 \over 2},P(X = j) = {1 \over {{3^j}}}(j = 1,2,3,...,\infty )$. Then the mean of the distribution and P(X is positive and even) respectively are :
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If the mean and variance of the following data : 6, 10, 7, 13, a, 12, b, 12 are 9 and ${{37} \over 4}$ respectively, then (a $-$ b)2 is equal to :
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The mean of the following probability distribution of a random variable $X$ is $\dfrac{46}{9}$.
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A person throws two fair dice. He wins Rs. $15$ for throwing a doublet (same numbers on the two dice), wins Rs. $12$ when the throw results in the sum of $9$, and loses Rs. $6$ for any other outcome on the throw. Then the expected gain/loss (in Rs.) of the person is:
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If a random variable $x$ has the probability distribution
then $P(3 < x \le 6)$ is equal to:
then $P(3 < x \le 6)$ is equal to:
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$\sum P(x) = 1$
$9k + 10k^2 = 1$
$\Rightarrow k = \frac{1}{10}$
$P(3 < x \le 6) = 3k + 3k^2$
$= \frac{3}{10} + \frac{3}{100} = 0.33$
A six faced die is biased such that
$3 \times P(\text{a prime number}) = 6 \times P(\text{a composite number}) = 2 \times P(1).$
Let $X$ be a random variable that counts the number of times one gets a perfect square on some throws of this die. If the die is thrown twice, then the mean of $X$ is:
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Let 9 distinct balls be distributed among 4 boxes, B1, B2, B3 and B4. If the probability than B3 contains exactly 3 balls is $k{\left( {{3 \over 4}} \right)^9}$ then k lies in the set :
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A random variable X has the following probability distribution :
The value of P(1 < X < 4 | X $\le$ 2) is equal to :
The value of P(1 < X < 4 | X $\le$ 2) is equal to :
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Let a random variable $X$ take values $0,1,2,3$ with $P(X=0)=P(X=1)=p$, $P(X=2)=P(X=3)$ and $E(X^2)=2E(X)$. Then the value of $8p-1$ is:
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The probability that the random variable $X$ takes value $x$ is given by
$P(X = x) = k(x + 1)3^{-x}, \; x = 0, 1, 2, 3, \ldots$
where $k$ is a constant. Then $P(X \ge 2)$ is equal to:
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