Probability of at Least 45 Heads in 100 Tosses of Fair CoinDate: 05/15/2004 at 08:14:21 From: Joe Subject: A different type of coin toss probability question What is the probability of getting AT LEAST 45 HEADS out of 100 tosses of a fair coin? I assumed that the answer would be P(45) + P(46) + .... + P(100). When I calculated that, it came out to .86437. However, an "expert" suggested the answer was P(45) + P(46) + .... + P(49) + P(50) + 0.5, where the 0.5 represents SUM P(51): P(100). Doing it that way, I came up with .90417. My concern is that if I were to "modify" the problem to "at least 41 heads", when I add the P(44) + P(43)+ P(42)+ P(41), I have already exceeded 100% probability with the .90417 answer. It leads me to believe that my original answer of .86437 might be correct. Please help by confirming if either is correct, or provide the correct answer if neither is correct. Thanks. Date: 05/15/2004 at 08:38:05 From: Doctor Douglas Subject: Re: A different type of coin toss probability question Hi Joe - Thanks for writing to the Math Forum. Your answer is correct. There is a subtle detail that slightly complicates the shorthand method. Since there are an EVEN number of coins, the number of possible outcomes is odd (e.g. 4 coins: 0,1,2,3,4 heads), and so the middle probability must be bisected at its halfway point: 0.5 = Pr(50)/2 + [Pr(51) + ... + Pr(100)] This explains the difference in your results: Pr(50)/2 = C(100,50)/2^101 = 100!/(50! x 50! x 2^101) = 0.0397946, equal to the discrepancy between the two numbers you calculated above. - Doctor Douglas, The Math Forum http://mathforum.org/dr.math/ Date: 05/16/2004 at 06:44:26 From: Joe Subject: Thank you (A different type of coin toss probability question) Dr. Douglas, Thanks for your prompt response and your explanation as to why the shortcut method does not work. It's great to have real experts when one needs help. Sincerely, Joe |
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