SOLUTION: find the probability of at least 3 successes in 6 trials of a binomial experiment in which the probability is 50%

Question 1137408: find the probability of at least 3 successes in 6 trials of a binomial experiment in which the probability is 50%
Answer by greenestamps(13215) (Show Source): You can put this solution on YOUR website!

The probability for success is 50% = 1/2, so the probability of failure is 1/2.

First method -- brute force application of binomial probability....

Second method -- using Pascal's Triangle....

Since the probabilities of success and failure are the same, the solution can be obtained with much less effort by knowing that the C(6,n) numbers are the numbers in the 6th row of Pascal's Triangle. It is a big time-saver in problems like this if you know (or can quickly compute) the numbers in the 6th row of Pascal's triangle. They are 1 6 15 20 15 6 1; the probability of at least 3 successes in 6 trials is then (20+15+6+1)/64 = 44/64 = 11/16.

Third method -- using more of what you know about Pascal's Triangle....

The pattern of numbers in each row of Pascal's Triangle is symmetrical. In the 6th row there are 7 numbers -- one in the middle, with identical sets of numbers on either side.

So you can compute C(6,3) = 20, giving you P(3 successes) = 20/64; then you know that the remaining 44/64 probability is equally for more than 3 successes or less than 3 successes. That makes P(more than 3 successes) = (44/64)/2 = 22/64; and that makes P(at least 3 successes) = 20/64+22/64 = 42/64 = 21/32.

Note that for this example, with 6 trials, the second method is faster than the third. However, for larger numbers of trials the third method might be faster.

Finally, note that this kind of problem is trivial if the number of trials is odd; for example, the probability of at least 8 successes in 15 trials is 1/2, because the 15th row of Pascal's Triangle contains 16 entries, with the first 8 entries being the same as the last 8 entries (in opposite order).
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