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We have a test of $10$ questions. Suppose we have a student that answer $8$ correct answers. What is the probability that he answers $4$ correct answers of the first $5$ questions? There isn't any connection between the questions in the test.

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As we already know that the student has answered correctly 8 questions, for him to answer $4$ correct answers in the $5$ first, he must have a mistake in the first $5$ ones, and another in the $5$ last ones. So there are, $5\times5=25$ ways to do this. The total number of ways to answer $8$ questions correctly in $10$ is $\binom{10}{8}=45$. Hence, the probability desired is $$P=\frac{25}{45}=\frac{5}{9}$$

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  • $\begingroup$ can you explain why ther are 25 ways to make 1 mistake in the first 5 and 1 mistake in the last 5? is it beacuse of (5 choose 4)*(5 choose 4)? $\endgroup$ – Aviad Chmelnik Dec 6 '13 at 14:03
  • $\begingroup$ You have $5$ ways to pick a quesiton to make a mistake in the first $5$ ones, and $5$ ways to pick in the $5$ last ones. So, $5\times 5=25$ $\endgroup$ – Mateus Sampaio Dec 6 '13 at 14:04
  • $\begingroup$ got it... thanks for the quick answer... $\endgroup$ – Aviad Chmelnik Dec 6 '13 at 14:05

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