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Probability Problem
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最佳解答:
This is a common negative binomial distribution question. Without any probability distribution knowledge, let us solve it in a high school way. The required probabilty is Pr( at least 5 times to get the second success of getting all the same ) First of all, note that Pr(getting all the same) = (1/2)3 × 2 = 1/4 = 0.25 (That is the case of HHH with prob. 1/8 and TTT with prob. 1/8) Then, Pr( at least 5 times to get the second success of getting all the same ) = Pr( within 4 times there is EXACTLY ONE success of getting all the same ) = Pr( within 4 times there is ONE success and THREE failures ) = ?C? (0.25) (1 - 0.25)3 = 4 × 0.25 × 0.753 = 0.753 = 27/64 = 0.421875 2014-04-29 12:48:00 補充： Sorry!! Both me and you were wrong. Please read again the supplementary revised answer. 2014-04-29 12:49:26 補充： I am sorry that the above calculation misses the cases where there is NO success case at all. The correct answer is 0.421875 (the above) + Pr(NO success within the 4 times) = 0.421875 + 0.75? = 0.421875 + 0.31640625 = 0.73828125 2014-04-29 12:52:14 補充： Your logic also works, but you just had a careless mistake. P(X < 5) = ?C? * (0.25)2 (0.75)? + ?C? * (0.25)2 (0.75)1 + ?C? (0.25)2 (0.75)2 = 0.0625 + 0.09375 + 0.10546875 = 0.26171875 P(X ≥ 5) = 1 - 0.26171875 = 0.73828125 [Ans] Basically, you are better than me. Excellent! (???)
其他解答:
My answer is different from yours. P(X >= 5) = 1 - P(X < 5) Consider the cases of getting 2nd success with less than 5 times of tossing: ? ? ? X ? ? X X ? P(X < 5) = 1C1 * (0.25)^1 (0.75)^0 + 2C1 * (0.25)^2 * (0.75) + 3C1 * (0.25)^2 * (0.75)^2 So P(X >= 5) = 0.5508 2014-04-29 12:40:09 補充： still waiting someone to clarify my answer whether it is correct or not.
Probability Problem
發問:
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John tosses three coins. Find the probability that he will have to toss at least five times to obtain the second successof getting all heads or all tails.最佳解答:
This is a common negative binomial distribution question. Without any probability distribution knowledge, let us solve it in a high school way. The required probabilty is Pr( at least 5 times to get the second success of getting all the same ) First of all, note that Pr(getting all the same) = (1/2)3 × 2 = 1/4 = 0.25 (That is the case of HHH with prob. 1/8 and TTT with prob. 1/8) Then, Pr( at least 5 times to get the second success of getting all the same ) = Pr( within 4 times there is EXACTLY ONE success of getting all the same ) = Pr( within 4 times there is ONE success and THREE failures ) = ?C? (0.25) (1 - 0.25)3 = 4 × 0.25 × 0.753 = 0.753 = 27/64 = 0.421875 2014-04-29 12:48:00 補充： Sorry!! Both me and you were wrong. Please read again the supplementary revised answer. 2014-04-29 12:49:26 補充： I am sorry that the above calculation misses the cases where there is NO success case at all. The correct answer is 0.421875 (the above) + Pr(NO success within the 4 times) = 0.421875 + 0.75? = 0.421875 + 0.31640625 = 0.73828125 2014-04-29 12:52:14 補充： Your logic also works, but you just had a careless mistake. P(X < 5) = ?C? * (0.25)2 (0.75)? + ?C? * (0.25)2 (0.75)1 + ?C? (0.25)2 (0.75)2 = 0.0625 + 0.09375 + 0.10546875 = 0.26171875 P(X ≥ 5) = 1 - 0.26171875 = 0.73828125 [Ans] Basically, you are better than me. Excellent! (???)
其他解答:
My answer is different from yours. P(X >= 5) = 1 - P(X < 5) Consider the cases of getting 2nd success with less than 5 times of tossing: ? ? ? X ? ? X X ? P(X < 5) = 1C1 * (0.25)^1 (0.75)^0 + 2C1 * (0.25)^2 * (0.75) + 3C1 * (0.25)^2 * (0.75)^2 So P(X >= 5) = 0.5508 2014-04-29 12:40:09 補充： still waiting someone to clarify my answer whether it is correct or not.
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