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CUET · MATHS · PYQ PAPER 2025

Let \(x\) denote the number of doublets in three throws of a pair of dice with the following probability distribution.
\(x\)\(0\)123
\(P(x)\)\(\frac{25}{72} k\)\(\frac{15}{72} k\)\(\frac{3}{72} k\)\(\frac{1}{360} k\)
If value of \(k\) is equal to \(\frac{m}{n}, \operatorname{gcd}(m, n)=1\), then \(m+n\) is equal to

  1. A 8
  2. B 19
  3. C 16
  4. D 18
Verified Solution

Answer & Solution

Correct Answer

(A) 8

Step-by-step Solution

Detailed explanation

\(\sum P(x) = 1\) \(\frac{25}{72}k + \frac{15}{72}k + \frac{3}{72}k + \frac{1}{360}k = 1\) \(\frac{43}{72}k + \frac{1}{360}k = 1\) \(\frac{43 \times 5}{360}k + \frac{1}{360}k = 1\) \(\frac{215}{360}k + \frac{1}{360}k = 1\) \(\frac{216}{360}k = 1\) \(k = \frac{360}{216}\)…
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