# Why does $10 \choose 7$ work in this case?

On a $10$ question true/false test, if the questions are answered at random, what is the probability of answering exactly $7$ questions correctly?

I see how $10 \choose 7$ would be choosing $7$ correct answers from $10$ possible answers, but how do I account for the $3$ false answers?

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Choosing $7$ correct answers corresponds to choosing $3$ incorrect answers. – The Chaz 2.0 Mar 23 '12 at 3:20
An answer is either true or false. "Choosing 7 correct answers" really means "choosing exactly 7 correct answers", so this accounts for "choosing exactly 3 incorrect answers" as well. – Matt Mar 23 '12 at 9:02

What you actually do is to choose $7$ correct answers from the $10$ and choose $3$ incorrect answers for the remaining $3$. Hence, we get $\binom{10}{7} \binom{3}{3} = \binom{10}{7}$.

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$\dbinom{10}{3}$ would be choosing 3 incorrect answers from the 10 possible answers. So the remaining 7 possible answers must be chosen correctly. In general, $\dbinom{n}{r} = \dbinom{n}{n-r}$.

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You want to accomplish three things:

1. Choose 7 questions. ($\binom{10}{7}$ ways)
2. Answer those 7 correctly. (probability $0.5$ for each)
3. Answer the remaining questions incorrectly. (probability $0.5$ for each)

Notice you don't need to pick the three incorrect answers. You are simply making sure that, once you've answered 7 correctly, all remaining questions are false.

Multiplying all that information together gives you a probability of $$\binom{10}{7}(0.5)^7(0.5)^3.$$

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