Question 1048494
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a)  There are 17 males in a class of 40, so the probability of selecting a male is *[tex \Large \frac{17}{40}].  Once a male has been selected, there are still 23 females to choose from, but now there are only 39 total students.  Hence, the probability of selecting a female on the second selection is *[tex \Large \frac{23}{39}].  Then once a male has been selected first and a female has been selected second, there are 22 females remaining out of a total class size of 38.  Probability of a female on the third selection is then *[tex \Large \frac{22}{38}\ =\ \frac{11}{19}].


Since these are three independent events (made independent based on adjusting the numerator and denominator for the individual probabilities of the three events), the total probability is the product of the three:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{17}{40}\ \times\ \frac{23}{39}\ \times\ \frac{11}{19}]


The rest is just arithmetic.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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