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Suppose a class of 50 students has 20 malee and 30 females. The instructor will pick one(different) student to compete in three different national competitions in math, chem, and english. What is the probability that there are at lrast two male students selected?

It is not really a case of solving. I know how to do that, I will take the complement thus: $$1 - [P(0 \\male) + P( 1\\ male) ]$$

it is more the counting I would like to know if I took the right approach. Let's use P(0 male) as the example. I interpreted it as such: $$\frac{\binom{30}{3} \binom{20}{0}}{\binom{50}{3}}$$

Now my rationale for this is that of there are to be 0 boys amd we need 3 people then they must come from females. Or should I interpret as if, if no males are to be selected then that is it amd thus: $$\frac{\binom{20}{0}}{\binom{50}{3}}$$

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  • $\begingroup$ The denominator in any case must be $\binom{50}{3}$ $\endgroup$ – true blue anil Dec 5 '15 at 17:27
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Both formulas are actually wrong for $0$ males, $3$ females.

The correct one is $\dfrac{\dbinom{30}{3}\dbinom{20}{0}}{\dbinom{50}{3}}$

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  • $\begingroup$ gosh darn it. the dang denominator......pardon me i am under the weather. So in terms of the numerator I should go ahead and assume that since 3 are required then they must be female then? $\endgroup$ – dc3rd Dec 5 '15 at 17:27
  • $\begingroup$ Yes, that's right. $\endgroup$ – true blue anil Dec 5 '15 at 17:43
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Let $\Pr(x)$ denote the probability that $x$ men are selected.

$$\Pr(0) = \frac{30\cdot 29 \cdot 28}{50\cdot 49 \cdot 48}$$ and $$\Pr(1) = \binom 31 \frac{30\cdot 29 \cdot 20}{50\cdot 49\cdot 48}.$$

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