W e know ∑Pi=1⇒6k2+5k=1
6k2+5k−1=0
6k2+6k−k−1=0
(6k−1)(k+1)=0⇒k=−1 (rejected) ; k=61
P(X>2)=k+2k+5k2
=61+62+365=366+12+5=3623
A random variable X has the following probability distribution:
X:12345P(X):k22kk2k5k2
Then, P(X>2) is equal to:
Held on 9 Jan 2020 · Verified 6 Jul 2026.
127
361
61
3623
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
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 :
Work through every JEE Main Probability & Statistics PYQ, year by year.