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$8$ students take an exam.

All of them are prepared average, so probability that they will pass or fail is the same.

After checking half of the tests, it's discovered that $3$ of them passed and $1$ failed.

What is probability that in the next $3$ tests, $1$ will pass and $2$ will fail

My reasoning:

If $A$ is event in which $1$ out of checked $4$ has passed and $2$ have failed and $B$ is event in which $3$ out of checked $4$ have passed and $1$ has failed then that two events are independent??

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The probability that each student passed or failed is independent of the other students. To see why imagine changing students passing a test with flipping a coin.

Since the probability of a student passing or failing is $.5$, we can just look at all possibilities for students passing or failing compared to the number of possibilities where 1 student passes and the other two fail.

Letting P = passing student, and F = failing student we have all possibilites are

PPP, PPF, PFP,FPP, FFP,FPF,PFF, FFF. There are $8$ total possibilities, and $3$ which satisfy our condition of $1$ student passing and $2$ students failing so we have a probability of $3/8$.

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