Probability & Statistics PYQ

From a month of $31$ days, $3$ different dates are selected at random. If the probability that these dates are in an increasing A.P. is equal to $\dfrac{a}{b}$, where $a,b \in \mathbb{N}$ and $\gcd(a,b)=1$, then $a+b$ 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 :

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 :

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

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 :

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 $\_\_\_\_$

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 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 :

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 :

A bag contains 10 balls out of which $k$ are red and ($10-k$) are black, where $0 \leq k \leq 10$. If three balls are drawn at random without replacement and all of them are found to be black, then the probability that the bag contains 1 red and 9 black balls is:

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:

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

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

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:

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:

The mean and variance of 10 observations are 9 and 34.2, respectively. If 8 of these observations are $2,3,5,10,11,13,15,21$, then the mean deviation about the median of all the 10 observations is

A bag contains 3 red and 5 blue balls. Two balls are drawn at random without replacement. The probability that both are red is:

The probabilities that players $A$ and $B$ of a team are selected for the captaincy for a tournament are $0.6$ and $0.4$, respectively. If $A$ is selected the captain, the probability that the team wins the tournament is $0.8$ and if $B$ is selected the captain, the probability that the team wins the tournament is $0.7$. Then the probability, that the team wins the tournament, is :

Let the mean and variance of 7 observations $2,4,10, x, 12,14, y, x>y$, be 8 and 16 respectively. Two numbers are chosen from $\{1,2,3, x-4, y, 5\}$ one after another without replacement, then the probability, that the smaller number among the two chosen numbers is less than 4, is :

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:

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