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GATE 2026 DA – Question 20

Probability and Statistics · Counting, probability axioms, conditional probability and Bayes theorem · 1 mark · Multiple choice

Suppose that a computer program provides a non-negative and integer-valued random solution to the equation $n_1 + n_2 + n_3 + n_4 = 20$.

Which of the following is the probability that all of $n_1, n_2, n_3, n_4$ in the provided solution are positive?

  1. $\binom{19}{3}\Big/\binom{23}{3}$
  2. $\binom{20}{4}\Big/\binom{24}{4}$
  3. $\binom{20}{3}\Big/\binom{23}{3}$
  4. $\binom{19}{4}\Big/\binom{24}{4}$

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Correct answer: (A) $\binom{19}{3}\Big/\binom{23}{3}$

Explanation

By stars and bars, the number of non-negative integer solutions of $n_1 + \cdots + n_4 = 20$ is $\binom{20 + 3}{3} = \binom{23}{3}$. The number of solutions with every $n_i \ge 1$ is $\binom{20 - 1}{3} = \binom{19}{3}$. With every solution equally likely, the probability is $\binom{19}{3}\Big/\binom{23}{3}$.