# Probability of getting $1$ out of $3$ shots

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"? –  Robert Israel Aug 17 '12 at 5:39
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%. –  Did Aug 17 '12 at 7:52
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. –  C. Williamson Aug 17 '12 at 17:40
@C.Williamson: The text is 1 out of 3, the rest is an interpretation by the OP. –  Did Aug 17 '12 at 22:43
@RobertIsrael its actually atleast –  MistyD Aug 18 '12 at 13:37
The way you do this is $3\choose{1}$(.8*.2*.2)+$3\choose{2}$(.8*.8*.2)+3\choose{3}$$(.8^3) This equals 0.992. There are 3\choose{1} ways to get one shot, 3\choose{2} ways to get two and 3\choose{3} ways to get all of them. Note that I could have just as easily counted misses rather than successful shots, which is okay, since binomial coefficients satisfy n\choose{k}=n\choose{n-k}. - add comment A way which nobody has mentioned is this: The probability is: getting the first throw + missing the first throw but getting the second + missing the first 2 but getting the third. 0.8+0.2*0.8+0.2*0.2*0.8=0.8+0.16+0.032=0.992 - add comment Assuming independence between the results of every two shots, then the scoring variable K out of n shots with success rate of p is (we write K~B(n,p))$$ Pr(K=k)=\binom{n}{k}p^k(1-p)^{n-k} $$(K=2 means two shots of success, and Pr() means probability.) So in your case:$$\begin{align} Pr(K>0)&=Pr(K=3)+Pr(K=2)+Pr(K=1) \\ &={3\choose 3}0.8^3 + {3\choose 2}0.8^2 0.2^1 + {3\choose 1}0.8^1 0.2^2 \\ &=0.512+0.384+0.096 \\ &=0.992 \\ &=1-Pr(K=0) \end{align}$So, actually$1-Pr(K=0)\$ would be the best way to go for -- easy calculation.