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- Mar 12th 2006, 10:20 AMJonPlease help
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- Mar 12th 2006, 04:12 PMThePerfectHackerQuote:

Originally Posted by**Jon**

He needs to be selected AND he would be the first. The probability that he is selected is $\displaystyle \frac{_5C_3\cdot _1C_1}{_6C_4}=\frac{2}{3}$

The probability that he is first is $\displaystyle 1/4$ thus, the answer is $\displaystyle \frac{1}{4}\cdot \frac{2}{3}=\frac{1}{6}$

Wishbone:

Same thing as in the first problem, same answer.

Bashful is first, Happy is last:First they need to be selected out of 6 people which is $\displaystyle \frac{_2C_2\cdot _4C_2}{_6C_4}=\frac{2}{5}$

Now bashful is first which is $\displaystyle 1/4$ and hapy is last which is $\displaystyle 1/3$ because only 3 poeple remain after bashful. Thus,

$\displaystyle \frac{2}{5}\cdot \frac{1}{4}\cdot \frac{1}{3}=\frac{1}{30}$

For the next problem, you need to select all 4 which is $\displaystyle \frac{_4C_4}{_6C_4}=\frac{1}{15}$, Now you need them is proper order which is for the first one $\displaystyle 1/4$ for the second $\displaystyle 1/3$ for the third $\displaystyle 1/2$ and for the last $\displaystyle 1/1$ Thus,

$\displaystyle \frac{1}{15}\cdot \frac{1}{4}\cdot \frac{1}{3}\cdot \frac{1}{2}\cdot \frac{1}{1}=\frac{1}{360}$

Sleepy Not last: Add that he is first or second or third. Which is $\displaystyle \frac{1}{6}$ as in the first problem. Thus, $\displaystyle \frac{1}{6}+\frac{1}{6}+\frac{1}{6}=\frac{1}{2}$

Wishbone not selected: If the probability of selecting him is $\displaystyle \frac{2}{3}$ then the probability of not selecting him is $\displaystyle \frac{1}{3}$

Hope this helps, and also that I did not erred. - Mar 12th 2006, 04:18 PMJonThanks
I did come up with some of the answers, but on others I see where I made my mistake. :cool: Thanks.

- Mar 12th 2006, 06:10 PMThePerfectHackerQuote:

Originally Posted by**Jon**