Question:medium
Let \( a \) and \( b \) be 2 consecutive integers selected from the first 20 natural numbers. The probability that \( \sqrt{a^2 + b^2 + a^2b^2} \) is an odd positive integer is
Let \( a \) and \( b \) be 2 consecutive integers selected from the first 20 natural numbers. The probability that \( \sqrt{a^2 + b^2 + a^2b^2} \) is an odd positive integer is
Show Hint
Try algebraic factorization to simplify complex expressions.
Updated On: May 1, 2026
- \( \frac{9}{19} \)
- \( \frac{10}{19} \)
- \( \frac{13}{19} \)
- \( 1 \)
- \( 0 \)
Show Solution
The Correct Option is A
Solution and Explanation
Was this answer helpful?
0
Top Questions on Probability
- The distribution function \( F(X) \) of a discrete random variable \( X \) is given. Then \( P[X = 4] + P[X = 5] \):
\[ \begin{array}{|c|c|c|c|c|c|c|} \hline X & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline F(X = x) & 0.2 & 0.37 & 0.48 & 0.62 & 0.85 & 1 \\ \hline \end{array} \]- MHT CET - 2024
- Mathematics
- Probability
- If \( X \) is a random variable with the probability mass function (p.m.f.) as follows: \[ P(X = x) = \begin{cases} \frac{5}{16}, & x = 0, \\ \frac{kx}{48}, & x = 1, \\ \frac{1}{4}, & x = 2, \\ \frac{1}{4}, & x = 3, \end{cases} \] then find \( E(X) \):
- MHT CET - 2024
- Mathematics
- Probability
- The p.m.f. of a random variable \( X \) is: \[ P(X) = \frac{2x}{n(n+1)}, \quad x = 1, 2, 3, \ldots, n \] \[ P(X) = 0, \quad \text{Otherwise.} \] Then \( E(X) \) is:
- MHT CET - 2024
- Mathematics
- Probability
- Find the expected value and variance of \( X \) for the following p.m.f: \[ \begin{array}{|c|c|c|c|c|c|} \hline x & -2 & -1 & 0 & 1 & 2 \\ \hline P(X) & 0.2 & 0.3 & 0.1 & 0.15 & 0.25 \\ \hline \end{array} \]
- MHT CET - 2024
- Mathematics
- Probability
- Want to practice more? Try solving extra ecology questions todayView All Questions
