JEE Main 2026 — Probability & Statistics

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:

A coin is tossed $8$ times. If the probability that exactly $4$ heads appear in the first six tosses and exactly $3$ heads appear in the last five tosses is $p$, then $96p$ is equal to _____.

A data consists of $20$ observations $x_1, x_2, \ldots, x_{20}$. If $\sum_{i=1}^{20}(x_i + 5)^2 = 2500$ and $\sum_{i=1}^{20}(x_i - 5)^2 = 100$, then the ratio of mean to standard deviation of this data is:

A letter is known to have arrived by post either from KANPUR or from ANANTPUR. On the envelope just two consecutive letters AN are visible. The probability, that the letter came from ANANTPUR, is:

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:

A random variable $X$ takes values $0,1,2,3$ with probabilities $\frac{2 a+1}{30}, \frac{8 a-1}{30}, \frac{4 a+1}{30}, b$ respectively, where $\mathrm{a}, \mathrm{b} \in \mathbf{R}$. Let $\mu$ and $\sigma$ respectively be the mean and standard deviation of X such that $\sigma^{2}+\mu^{2}=2$. Then $\frac{\mathrm{a}}{\mathrm{b}}$ is equal to :

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:

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 :

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

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

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 ______

From the first 100 natural numbers, two numbers first $a$ and then $b$ are selected randomly without replacement. If the probability that $a-b \geqslant 10$ is $\frac{m}{n}, \operatorname{gcd}(m, n)=1$, then $m+n$ is equal to $\_\_\_\_$.

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 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

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 :

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 :

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

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

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 $\_\_\_\_$

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 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 :

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 :

Suppose that the mean and median of the non-negative numbers $21, 8, 17, a, 51, 103, b, 13, 67, (a > b)$, are $40$ and $21$, respectively. If the mean deviation about the median is $26$, then $2a$ 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

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

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:

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 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 :

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 :

Two distinct numbers $a$ and $b$ are selected at random from $1,2,3, \ldots, 50$. The probability, that their product $a b$ is divisible by 3, is