The expected value of the sum of the two numbers obtained on the uppermost faces,when two fair dice are rolled,is

  • A
    $7$
  • B
    $12$
  • C
    $6$
  • D
    $5$

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$A$ random variable $X$ takes values $-1, 0, 1, 2$ with probabilities $\frac{1+3p}{4}, \frac{1-p}{4}, \frac{1+2p}{4}, \frac{1-4p}{4}$ respectively,where $p$ varies over $\mathbb{R}$. Then the minimum and maximum values of the mean of $X$ are respectively.

$A$ fair die is tossed twice in succession. If $X$ denotes the number of sixes in two tosses,then the probability distribution of $X$ is given by

The probability distribution of a random variable $X$ is given below.
$X = x$ $0$ $1$ $2$ $3$
$P(X = x)$ $\frac{1}{10}$ $\frac{2}{10}$ $\frac{3}{10}$ $\frac{4}{10}$

Then the variance of $X$ is

Which of the following can not be a valid assignment of probabilities for outcomes of sample space $S = \{\omega_{1}, \omega_{2}, \omega_{3}, \omega_{4}, \omega_{5}, \omega_{6}, \omega_{7}\}$?
OutcomeProbability
$\omega_{1}$$0.1$
$\omega_{2}$$0.01$
$\omega_{3}$$0.05$
$\omega_{4}$$0.03$
$\omega_{5}$$0.01$
$\omega_{6}$$0.2$
$\omega_{7}$$0.6$

$A$ fair die is tossed repeatedly until a six is obtained. Let $X$ denote the number of tosses required and let $a=P(X=3)$,$b=P(X \geq 3)$ and $c=P(X \geq 6 \mid X>3)$. Then $\frac{b+c}{a}$ is equal to

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