The mean and standard deviation of the marks of 10 students were found to be 50 and 12 respectively. Later, it was observed that two marks 20 and 25 were wrongly read as 45 and 50 respectively. Then the correct variance is
Held on 13 Apr 2023 · Verified 6 Jul 2026.
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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:
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
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:
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 ______
Work through every JEE Main Probability & Statistics PYQ, year by year.