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The question is:

Probability of scoring $1$ out of $3$ shots when you have 80% throw rate.

I solved this problem in the inverse way:

P(At least one basket) = $1$ - P(No basket) = $1$ - ($.20 \times .20 \times .20) = 0.992 ~ or ~99.2 \% $

I wanted to know how would I go about solving this the other (which I realize is not the best way)..

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    Which do you mean, "at least one" or "exactly one"?2012-08-17
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    If I had to put some money on the exact meaning of *Probability of scoring 1 out of 3 shots*, I would bet on *exactly one hit*. Then the numerical answer is 9.6%.2012-08-17
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    I agree that this is confusing, but technically, if you score twice, you have scored once. Also, the questioner supports this via his/her calculations. She also mentions P(At least one basket), which isn't very ambiguous.2012-08-17
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    @C.Williamson: The text is **1 out of 3**, the rest is an interpretation by the OP.2012-08-17
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    @RobertIsrael its actually atleast2012-08-18

3 Answers 3