Choosing the correct combinatorial method

A box contains 24 lightbulbs 2 of which are defective. If a person selects 10 lightbulbs at random without replacement, what is the probability that both defective bulbs will be selected?

I'm searching for the right way to select the sample space.

The denominator should be all the possible ways in which 10 balls can be selected from 24, i.e. ${24\choose 10}$

and the numerator should be all the possible ways in which the defective balls can be selected, but I can't decide whether that is: $2^{10}$ or just simply ${10\choose 2}$ can someone help clarify this?

EDIT:

second question:

Suppose that 35 people are divided in a random manner into two teams in such a way that one team contains 10 people and the other team contains 25 people. What is the probability that two particular people A and B will be on the same team?

i'm still having trouble finding the event space. The sample space should be all the possible ways in which 10 and 25 can be selected from 35, i.e. (35 choose 10)*(35 choose 10). But I have no idea how to find the event space... it seems like there are only 2 possibilities but I guess I'm wrong..

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You need to ask separate questions as separate questions. – Zev Chonoles Oct 16 '11 at 22:07

To be precise, the size of your sample space is $\binom{24}{10}$. This number does go on the bottom of the fraction, and what goes on top is the size of the event. Break up the event into independent events 1. choose the 2 defective bulbs, and 2. choose the remaining 8 bulbs. I don't have much choice in event 1. There is only one way to choose both of the defective balls. In other words, $\binom{2}{2}$ (choosing 2 defective bulbs from a set of 2 defective bulbs). For event 2, there are $24-2 = 22$ nondefective bulbs, and I must choose $8$ of them, so that's $\binom{22}{8}$. Finally, since events 1 and 2 are independent, we multiply the answers for the combined event: $\binom{2}{2}\binom{22}{8}$

$$P = \frac{\binom{2}{2}\binom{22}{8}}{\binom{24}{10}}$$

Or, since $\binom{2}{2} = 1$,

$$P = \frac{\binom{22}{8}}{\binom{24}{10}}$$

Hope this helps!

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it definitely helps. thanks! – Dbr Oct 16 '11 at 21:23
please see my edited question. sorry for asking but I'm trying to understand how to do this.. – Dbr Oct 16 '11 at 21:34
Your second question is unrelated to the first; you should start a new question. – sasha Oct 16 '11 at 21:56

To get the numerator, you need to find the number of ways to select the $2$ defective bulbs and $8$ of the remaining $22$ bulbs. How many ways are there to choose $8$ bulbs from a set of $22$ bulbs? Once you’ve done that, you have no further choices: you have to pick the two defective bulbs to fill out the set of $10$.

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For the second question, the team of $10$ can be chosen in $\binom{35}{10}$ ways, all equally likely. Once we have done the choosing of the team of $10$, we are done. So a suitable sample space is all the possible teams of $10$.

In how many of these choices are A and B on the same team? This can happen in two different ways: (i) A and B are on the team of $10$ and (ii) A and B are on the team of $25$, or to put it another way, are not on the team of $10$.

(i) We count the number of ways to choose the team of $10$ so that A and B are both on the team. So we must choose $8$ people from the $33$ who are not A or B, to be teammates of A and B. This can be done in $\binom{33}{8}$ ways.

(ii) We count the number of ways to choose the team of $10$ so that neither A nor B is on the team. So we must choose $10$ people from $33$, and this can be done in $\binom{33}{10}$ ways.

We conclude that the number of ways to choose the team of $10$ so that A and B are both on that team, or both on the other team, is $$\binom{33}{8}+\binom{33}{10}.$$ So the probability that A and B are on the same team is $$\frac{\binom{33}{8}+\binom{33}{10}}{\binom{35}{10}}.$$

When we calculate, we can arrange things so that there is considerable simplification.

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Another way for the $2nd$ question is to use direct probabilities of both being in groups of 10 or 25

$$\frac{10}{35}\cdot\frac{9}{34} + \frac{25}{35}\cdot\frac{24}{34}$$

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