Question 2

# Question 2 - Therefore, we can write: But now, the P(X 1 +...

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Question 2 Fifteen flips of a biased coin are conducted in which the probability of a head is 0.6. Please calculate/provide exact probabilities for each answer. Let X 1 = the number of heads in the first ten flips. Let X 2 = the number of heads in the last five flips. Let X 3 = the number of heads in the last (fifteenth) flip. c.) Find P(X 1 +X 2 +X 3 10) What are the domains of X 1 , X 2 , X 3 ? Therefore, the domain of . This doesn’t have much bearing on the solution, but it is good to keep in mind. So we want to find the P(Y<=10). In order to do this we need to investigate 2 scenarios, when X 3 = 0 and when X 3 = 1. Why? If we know X 3 = 0, then we must have 10 or less heads in the first 14 flips. Else, if we know X 3 = 1, then we must have 9 or less heads in the first 14 flips.
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Unformatted text preview: Therefore, we can write: But now, the P(X 1 + X 2 <=10) is distributed as a Binomial(N=14,p=.6), the cumulative probability of 10 heads or less in the first 14 flips. This is because we are told that the 15 th flip is not a head and can therefore have at most 10 heads in the first 14 flips. Also, the P(X 1 + X 2 <=9) is distributed similarly as a Binomial(N=14,p=.6) and we are looking for the cumulative probability of 9 heads in the first 14 flips. This is because we are told that the 15 th flip is a head and therefore can have at most 9 heads in the first 14 flips. = BINOMDIST(10,14,.6,true)(.4) + BINOMDIST(9,14,.6,true)(.6) And from the cumulative binomial tables we know: =(.876)(.4)+(.721)(.6)...
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## This note was uploaded on 01/04/2010 for the course STATS 1100 taught by Professor Rodriguez during the Spring '08 term at Pittsburgh.

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