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Suppose a box contains 5 white balls and 5 black balls.

If you want to extract a ball and then another:

What is the probability of getting a black ball and then a black one?

I think that this is the answer:

Let $A:$ get a black ball in the first extraction, $B:$ get a black ball in the second extraction. Then: $P(A)=\frac{5}{10}$ and $$P(B|A)=\frac{P(A\cap B)}{P(A)}=\frac{\frac{4}{9}}{\frac{1}{2}}=\frac{8}{9}$$

Is this correct?

Now, If you want to extract two balls at the same time: what is the probability that both are black? Is exactly the same probability of the first question? Why?

Thanks for your help.

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1  
Regarding the first question, is the first ball replaced after extraction or not? –  wnvl Sep 19 '12 at 16:22
1  
By symmetry, you expect the answer to be the same if you replace "black" by "white". Do you see a problem with 8/9 being the answer then? –  mrf Sep 19 '12 at 16:24
    
@wnvl No, the ball is not replaced. –  Hiperion Sep 19 '12 at 16:25
    
According to the text, what you want to calculate is the probability of getting two black balls if you draw two balls one after the other ("a black one and then a black one"). According to your used formula and notation, what you want to calculate is the probability that the second drawn ball is black if the first one was black. So which one do you actually want to calculate? –  celtschk Sep 19 '12 at 16:28
    
@mrf No, I don't think so, because the answer is too "big". –  Hiperion Sep 19 '12 at 16:33

4 Answers 4

up vote 4 down vote accepted

It is good to check an answer for plausibility. Given that the first ball is black, surely the probability that the second ball is black cannot be $\dfrac{8}{9}$!

In this case, it is the conditional probability that is obvious. Note that $\Pr(B|A)=\dfrac{4}{9}$. And what we want is $\Pr(A\cap B)$. We get $$\Pr(A\cap B)=\Pr(B|A)\Pr(A)=\frac{4}{9}\cdot\frac{1}{2}.$$

As to your question about two consecutive picks versus two simultaneous picks, there is no effective difference. If Alicia picks two balls from the urn, what difference can it make whether she touches the two balls simultaneously, or with a delay of a millisecond?

Remark: For this kind of problem, using the formal conditional probability machinery is overkill, and may hinder the intuition.

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Thnak you very much! :) –  Hiperion Sep 19 '12 at 16:53

If the first ball is not replaced after extraction, both probabilities are the same.

$P(\text{first ball black and second ball black})$

$=P(\text{first ball black}) \cdot P(\text{second ball black } | \text{ first ball black})$

$=\frac{5}{10}\cdot\frac{4}{9}$

$=\frac{2}{9}$

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+1 Thank you very much. –  Hiperion Sep 19 '12 at 16:58

Not quite. Now, if the first ball isn't replaced before drawing the second, then $P(B|A)=\frac49$, from which $$P(A\cap B)=P(A)\cdot P(B|A)=\frac12\cdot\frac49=\frac29.$$ If we are replacing the first ball, $P(B|A)=P(B)=\frac12$, and that changes things.

As for the second question, let's use combinatorics. There are $5$ black balls, and we want to choose $2;$ $5$ white balls, choose $0$. Total, there are $10$ balls, and we're choosing $2$. Then the probability of choosing $2$ black balls simultaneously is $$\frac{_5C_2\cdot{}_5C_0}{_{10}C_2}=\frac{10\cdot 1}{45}=\frac29.$$ Thus, we have the same result (if I'm right about not replacing the first ball, anyway).

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+1 Yes, you are right about not replacing the first ball. Thank you very much –  Hiperion Sep 19 '12 at 16:55

Why would you even split it into two events?!

You are just picking two balls. There are C(10,2)=45 ways to pick two balls out of 10, but only C(5,2)=10 give you two black ones. So 10/45 = 2/9 is your answer.

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