Probability & Statistics PYQ

The probability distribution of a random variable $X$ is given below : \(\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline X & 4k & \dfrac{30}{7}k & \dfrac{32}{7}k & \dfrac{34}{7}k & \dfrac{36}{7}k & \dfrac{38}{7}k & \dfrac{40}{7}k & 6k \\ \hline P(X) & \dfrac{2}{15} & \dfrac{1}{15} & \dfrac{2}{15} & \dfrac{1}{5} & \dfrac{1}{15} & \dfrac{2}{15} & \dfrac{1}{5} & \dfrac{1}{15} \\ \hline \end{array}\) If $\mathrm{E}(\mathrm{X})=\frac{263}{15}$, then $\mathrm{P}(\mathrm{X}<20)$ is equal to :

If X follows a Poisson distribution with P(X=1) = P(X=2) then the mean of the distribution is:

The mean deviation about the mean for the data <table class="pyq-table"><tbody><tr><td>$x_i$</td><td>$5$</td><td>$7$</td><td>$9$</td><td>$10$</td><td>$12$</td><td>$15$</td></tr><tr><td>$f_i$</td><td>$8$</td><td>$6$</td><td>$2$</td><td>$2$</td><td>$2$</td><td>$6$</td></tr></tbody></table>is equal to:

A man throws a fair coin repeatedly. He gets $10$ points for each head he throws and $5$ points for each tail he throws. If the probability that he gets exactly $30$ points is $\dfrac{m}{n}$, $\gcd(m, n) = 1$, then $m + n$ is equal to:

If the mean deviation about the median of the numbers $\mathrm{k}, 2 \mathrm{k}, 3 \mathrm{k}, \ldots. ., 1000 \mathrm{k}$ is 500, then $\mathrm{k}^{2}$ is equal to :

From a lot containing 10 defective and 90 non-defective bulbs, 8 bulbs are selected one by one with replacement. Then the probability of getting at least 7 defective bulbs is

A bag contains $(N+1)$ coins $- N$ fair coins, and one coin with 'Head' on both sides. A coin is selected at random and tossed. If the probability of getting 'Head' is $\dfrac{9}{16}$, then $N$ is equal to:

Let the mean and the variance of seven observations $2, 4, \alpha, 8, \beta, 12, 14$, $\alpha < \beta$, be $8$ and $16$ respectively. Then the quadratic equation whose roots are $3\alpha + 2$ and $2\beta + 1$ is :

If a random variable $x$ has the probability distribution $\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline P(x) & 0 & 2k & k & 3k & 2k^{2} & 2k & k^{2}+k & 7k^{2} \\ \hline \end{array}$ then $P(3 < x \leq 6)$ is equal to

If the mean of the data <table class="pyq-table"><tbody><tr><th>Class</th><th>$5-10$</th><th>$10-15$</th><th>$15-20$</th><th>$20-25$</th><th>$25-30$</th><th>$30-35$</th></tr><tr><td>Frequency</td><td>$2$</td><td>$k$</td><td>$28$</td><td>$54$</td><td>$k+1$</td><td>$5$</td></tr></tbody></table> is $21$, then $k$ is one of the roots of the equation :

If the mean and the variance of the data \(\begin{array}{|c|c|c|c|c|} \hline \text{Class} & 4\text{-}8 & 8\text{-}12 & 12\text{-}16 & 16\text{-}20 \\ \hline \text{Frequency} & 3 & \lambda & 4 & 7 \\ \hline \end{array}\) are $\mu$ and 19 respectively, then the value of $\lambda+\mu$ is

A bag contains 6 blue and 6 green balls. Pairs of balls are drawn without replacement until the bag is empty. The probability that each drawn pair consists of one blue and one green ball is :

A variable $X$ takes values $0, 0, 2, 6, 12, 20, \ldots, n(n-1)$ with frequencies ${}^nC_0, {}^nC_1, {}^nC_2, {}^nC_3, {}^nC_4, {}^nC_5, \ldots, {}^nC_n$, respectively. If the mean of this data is $60$, then its median is :

Let the mean and variance of 8 numbers $-10,-7,-1, x, y, 9,2,16$ be $\frac{7}{2}$ and $\frac{293}{4}$, respectively. Then the mean of 4 numbers $x, y, x+y+1,|x-y|$ is :

The mean and variance of a data of 10 observations are 10 and 2, respectively. If an observations $\alpha$ in this data is replaced by $\beta$, then the mean and variance become 10.1 and 1.99, respectively. Then $\alpha+\beta$ equals

A set of four observations has mean $1$ and variance $13$. Another set of six observations has mean $2$ and variance $1$. Then, the variance of all these $10$ observations is equal to:

The mean and variance of $n$ observations are $8$ and $16$, respectively. If the sum of the first $(n-1)$ observations is $48$ and the sum of squares of the first $(n-1)$ observations is $496$, then the value of $n$ is:

Let S be a set of 5 elements and $\mathrm{P}(\mathrm{S})$ denote the power set of S. Let E be an event of choosing an ordered pair ($\mathrm{A}, \mathrm{B}$) from the set $\mathrm{P}(\mathrm{S}) \times \mathrm{P}(\mathrm{S})$ such that $\mathrm{A} \cap \mathrm{B}=\emptyset$. If the probability of the event $E$ is $\frac{3^{p}}{2^{q}}$, where $p, q \in N$, then $p+q$ is equal to $\_\_\_\_$

For $10$ observations $x_1, x_2, \ldots, x_{10}$, if $\sum_{i=1}^{10}(x_i+2)^2=180$ and $\sum_{i=1}^{10}(x_i-1)^2=90$, then their standard deviation is:

Let $\mathrm{X}=\{x \in \mathrm{~N}: 1 \leq x \leq 19\}$ and for some $a, b \in \mathbb{R}, \mathrm{Y}=\{a x+b: x \in \mathrm{X}\}$. If the mean and variance of the elements of Y are 30 and 750, respectively, then the sum of all possible values of $b$ is

A candidate has to go to the examination centre to appear in an examination. The candidate uses only one means of transportation for the entire distance out of bus, scooter and car. The probabilities of the candidate going by bus, scooter and car, respectively, are $\dfrac{2}{5}$, $\dfrac{1}{5}$ and $\dfrac{2}{5}$. The probabilities that the candidate reaches late at the examination centre are $\dfrac{1}{5}$, $\dfrac{1}{3}$ and $\dfrac{1}{4}$ if the candidate uses bus, scooter and car, respectively. Given that the candidate reached late at the examination centre, the probability that the candidate travelled by bus is:

Bag A contains 9 white and 8 black balls, while bag B contains 6 white and 4 black balls. One ball is randomly picked up from the bag B and mixed up with the balls in the bag A. Then a ball is randomly drawn from the bag A. If the probability, that the ball drawn is white, is $\frac{\mathrm{p}}{\mathrm{q}}, \operatorname{gcd}(\mathrm{p}, \mathrm{q})=1$, then $\mathrm{p}+\mathrm{q}$ is equal to