InterviewSolution
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InterviewSolution
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Two players `P_1`, and `P_2`, are playing the final of a chase championship, which consists of a series of match Probability of `P_1`, winning a match is 2/3 and that of `P_2` is 1/3. The winner will be the one who is ahead by 2 games as compared to the other player and wins at least 6 games. Now, if the player `P_2`, wins the first four matches find the probability of `P_1`, wining the championship. |
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Answer» `P_(1)` can win in the following matually exclusive ways: (a) `P_(1)` wins the next dix matches. (b) `P_(1)` wins five out of next six matches, so that after new six matches scored of `P_(1)and P_(2)` are tied up. This tie continues up to next 2n matches `(nge0)` and finally `P_(1)` wins 2 consective matches. Now, for case (a), probability is given by `(2//3)^(6)` and probability of tie after 6 matches [in case(b)] is `""^(6)C_(5)((2)/(3))^(5)((1)/(3))=6xx(2^(5))/(3^(6))=(2^(6))/(3^(5))` Now probability that scores are still tied up after another next two matches is `2/3xx1/3+1/3xx2/3=4/9` [First match won by `P_(1)` and second by `P_(2)` or first by `P_(2)` and second by `P_(1).]` Similarly, probability that scores are still tied up after another 2n matches is `(4//9)^(n).` Therefore, the total probability of `P_(1)` winning the championship is `((2)/(3))^(6)+(2^(6))/(3^(5))(underset(n=0)overset(oo)sum((4)/(9))^(n)((2)/(3))^(2))` `=((2)/(3))^(6)+(2^(6))/(3^(5))((2)/(3))^(2)((1)/(1-(4)/(9)))` `=17/5((2)/(3))^(6)=1088/3645` |
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