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A company is hiring candidates for four CEO positions. Five candidates are women, and three are men.

Given that all eight candidates are equally qualified, and that every combination of male and female candidates is equally likely to be chosen, what is the probability that at least one man will be chosen?

NB- Please give your answer as a fraction, e.g. 1/2, not 0.5

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    Any thoughts? Hint: it's easier to compute the probability that all $4$ choices are women. – 2017-01-28
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    A special gift for the first person that gets it right – 2017-01-28
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    1 - (probability that they are all women) – 2017-01-28
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    Because the Q is closed, I'll try to squeeze another kind of solution into a Comment: You may want to consider the number $X$ of men chosen as a hypergeometric random variable. You seek $P(X \ge 1).$ In R statistical software, the PDF is `dhyper`. Then `sum(phyper(1:4), 3, 5, 4)` returns 0.9285714 = 13/14. You also get the answer from `1 - dhyper(0, 3 ,5, 4)`. The hypergeometric PDF can be expressed in terms of binomial coefficients. (Look it up, the messiest part is the name 'hypergeometric'.) – 2017-01-29

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We can use complementary counting. Calculate the probability that none of the positions are held by men:

$$\frac 58 \times \frac 47 \times \frac 36 \times \frac 25=\frac 1{14}$$

Then you subtract that probability from $1$, giving you your answer of $\dfrac {13}{14}$

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    13/14 is correct. – 2017-01-28
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    $\!\!\!\!$ ok... – 2017-01-28
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    Are you testing us or asking for help? – 2017-01-28
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    Asking for help. Thanks – 2017-01-28