Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2 white balls. One ball is transferred from Bag I to Bag II and then a ball is draw from Bag II. The ball so drawn is found to be black in colour. Then the probability, that the transferred ball is red, is
Held on 29 Jul 2022 · Verified 6 Jul 2026.
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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:
If X follows a Poisson distribution with P(X=1) = P(X=2) then the mean of the distribution is:
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
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 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 :
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