Q. Two people, P and Q, decide to independently roll two identical dice, each with 6 faces, numbered 1 to 6. 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 P and Q. Assume that all 6 numbers on each dice are equi-probable and that all trials are independent. The probability (rounded to 3 decimal places) that one of them wins on the third trial is .<\/p>\n\n\n\n
Ans: 0.023<\/p>\n\n\n\n
Sol:<\/p>\n\n\n\n
Given there are two identical dices, each having 6 faces.
Favorable events for tie = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}
Since two dice are thrown, total sample space will be 6 x 6 = 36<\/p>\n\n\n\n
Therefore, P(tie) = 6\/36 = 1\/6
and Probability of not tie = (1 \u2013 1\/6)<\/p>\n\n\n\n
To one of them win on third trial, previous two trials should be tie.<\/p>\n\n\n\n
= 1\/6 * 1\/6 * (1 \u2013 1\/6)
= 1\/36 * 5\/6
= 5\/216 <\/p>\n\n\n\n
= 0.023<\/p>\n","protected":false},"excerpt":{"rendered":"
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