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Official PYQ · 2018Q15Probability: Random Variables and DistributionsNAT1 mark

GATE CS 2018 Q15 — Probability: Random Variables and Distributions

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Two people, PP and QQ, decide to independently roll two identical dice, each with 66 faces, numbered 11 to 66. The person with the lower number wins. In case of a tie, they roll the dice repeatedly until there is no tie. Define a trial as a throw of the dice by PP and Q.Q. Assume that all 66 numbers on each dice are equi-probable and that all trials are independent. The probability (rounded to 33 decimal places) that one of them wins on the third trial is ____
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Answer: 0.0230 : 0.0232

Let PP and QQ roll their respective fair 6-sided dice in each trial.

For any single trial, the total number of possible outcomes is 6imes6=366 imes 6 = 36.

Step 1: Probability of a Tie

A tie occurs when both PP and QQ roll the same number. The favorable outcomes for a tie are:

{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}\{(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)\}

Thus, the probability of a tie in any single trial is:

ptie=636=16p_{\text{tie}} = \frac{6}{36} = \frac{1}{6}

Step 2: Probability of a Win

A win occurs for either player when there is no tie (i.e., one of them rolls a lower number than the other). The probability of a win in any single trial is:

pwin=1−ptie=1−16=56p_{\text{win}} = 1 - p_{\text{tie}} = 1 - \frac{1}{6} = \frac{5}{6}

Step 3: Probability of Winning on the Third Trial

For the game to be decided on the third trial, the first two trials must result in ties, and the third trial must result in a win. Since each trial is independent, we have:

Probability=ptie×ptie×pwin\text{Probability} = p_{\text{tie}} \times p_{\text{tie}} \times p_{\text{win}} Probability=16×16×56=5216\text{Probability} = \frac{1}{6} \times \frac{1}{6} \times \frac{5}{6} = \frac{5}{216}

Step 4: Decimal Calculation

5216≈0.023148\frac{5}{216} \approx 0.023148

Rounding to three decimal places, the probability is 0.023 (which lies in the range 0.0230 : 0.0232).

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