SOLUTION: 1. Suppose we want to determine the (binomial) probability (p) of getting 6 heads in 13 flips of a 2-sided coin. Using Table 2 in Appendix B of the text, what values of n, x, and

Algebra ->  Algebra  -> Probability-and-statistics -> SOLUTION: 1. Suppose we want to determine the (binomial) probability (p) of getting 6 heads in 13 flips of a 2-sided coin. Using Table 2 in Appendix B of the text, what values of n, x, and       Log On

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 Algebra: Probability and statistics Solvers Lessons Answers archive Quiz In Depth

 Question 511397: 1. Suppose we want to determine the (binomial) probability (p) of getting 6 heads in 13 flips of a 2-sided coin. Using Table 2 in Appendix B of the text, what values of n, x, and p would we use to look up this probability, and what would be the probability? This is what I think I know: n = 11 x = 6 p = .50q = 1-p = .50 I think this is the forumula I use: 11/(11-6)6 (.50)^2(.50)^5 = I can't figure out any more or even if what I have done is correct. I need some line for line, number to number detail so I can really learn and understand this please.Answer by stanbon(57347)   (Show Source): You can put this solution on YOUR website! Suppose we want to determine the (binomial) probability (p) of getting 6 heads in 13 flips of a 2-sided coin. --- The 6 heads can show up is 13C6 = 1716 mutually exclusive ways. --- Each of those ways has probability (1/2)^6*(1/2)^7 = 1/2^13 = 1/8192 ==== Because the 1716 ways to succeed are mutually exclusive you should add all 1716 of those separate probabilities to get the final probability. ------ You could add but it's easier to just multiply 1716*(1/8192) = 0.2095 --- In general, if there are n trials, and the probability of success on each trial is "p", the probability of "k" successes is as follows: P(x = k) = nCk*p^k*q^(n-k) ============================== Cheers, stan H. ===============