Suppose that at the shooting range you take 10 shots at the target, with a .75 p

Question

Suppose that at the shooting range you take 10 shots at the target, with a .75 probability on each shot of hitting the target. Assume that each shot is independent of all the others. To find the probability of exactly 7 hits and 3 misses (in any order), the formula requires us to compute C(10, 7)(.75)^7 (.25)^3 a. If we just computed (.75)^7 (.25)^3 without multiplying by C(10.7), of what event would that be the probability? Why must multiply (.75)^7 (.25)^3 by C(10, 7) when we answer the original question about 7 hits and 3 misses? What, specifically, does the factor of C(10, 7) account for?

Explanation / Answer

a. It means exactly 7 successes and 3 failures in 10 trials.

b. When one computes exactly 7 succeses and 3 failures, then it implies that one shots target in first 7 trials but not in the next 3-in that order. But, one can shots target in 2nd, 4th, 6th, 7th, 8th, 9th and 10th trials and still have 7 succeses. The probability would be 0.25*0.75*0.25*0.75*0.25*0.75*0.75*0.75*0.75, and that is also (0.75)^7(0.25)^3. Fortunately these possible orders are disjoint. For small number of trials, it can be computed manually but for large n, one need to use formula to compute k succeses in n trials. That is why the factor C(10,7) is multiplied with (0.75)^7(0.25)^3.

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