x+y=24,x,y∈NAM>GM⇒xy≤144xy≥108 Favorable pairs of (x,y) are (13,11),(12,12),(14,10),(15,9),(16,8),(17,7),(18,6),(6,18),(7,17),(8,16),(9,15),(10,14),(11,13) i.e. 13 cases
Total choices for x+y=24 is 23 Probability =2313=nmn−m=10
Let the sum of two positive integers be 24 . If the probability, that their product is not less than 43 times their greatest possible product, is nm, where gcd(m,n)=1, then n−m equals
Held on 8 Apr 2024 · Verified 6 Jul 2026.
10
9
11
8
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.