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From A First Course in Probability (9th Edition):

3.5 An urn contains 6 white and 9 black balls. If 4 balls are to be randomly selected without replacement, what is the probability that the first 2 selected are white and the last 2 black?

This method is straightforward and results in the correct answer (according to the book): $$\frac{6}{15} \cdot \frac{5}{14} \cdot \frac{9}{13} \cdot \frac{8}{12} = \frac{6}{91} $$

(This is just the multiplication principle and probability of drawing the color of that ball at that time)

However, I want to understand this in terms of conditional probability. I don't understand why this doesn't work:

$$P(E \mid F) = \frac{P(E \cap F)}{P(F)} ={\frac{{6 \choose{2}}{9 \choose 2}}{{15 \choose{2}}{13 \choose 2}}}÷{\frac{{6 \choose{2}}}{{15 \choose{2}}}} = {\frac{{9 \choose 2}}{{13 \choose 2}}} = \frac{6}{13} \ne \frac{6}{91}$$

$\frac{6}{13}$ is exactly 7 times more than the previous answer. Why does this method fail to work? What mistake have I made? I tried to use the exact same method used in question 3.3, where this resulted in the correct answer.

Optional – About 3.3

3.3 Use Equation (2.1) to compute in a hand of bridge the conditional 
probability that East has 3 spades given taht North and South have a 
combined total of 8 spades.

Here, we see that: $$P(E \mid F) = \frac{P(E \cap F)}{P(F)} ={\frac{{13 \choose{8}}{39 \choose 18}{5 \choose 3}{21 \choose 10}} {{52 \choose{26}}{26 \choose 13}}}÷{\frac{{13 \choose{8}}{39 \choose 18}}{{52 \choose{26}}}} = {\frac{{5 \choose 3}{21 \choose 10}}{{26 \choose 13}}} = \frac{29}{115} \approx 0.339$$

Which is the answer in the back of the book.

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    There is no "condition" here. The reason your answer is exactly $7$ times the correct answer is because you divided by $P(F)$ which is $\frac 1 7$. – Noble Mushtak Feb 07 '16 at 15:22
  • @NobleMushtak I think you just made me get it. 3.3 asks for East's hand GIVEN the other hands, while this one asks it GENERALLY. Is that correct? Otherwise, they are almost the same. –  Feb 07 '16 at 15:24
  • Maybe I am too tired to be doing this now, because if that is it, I have to admit – I feel a bit dumb! xD –  Feb 07 '16 at 15:24
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    Yes, that's correct. – Noble Mushtak Feb 07 '16 at 15:25
  • @NobleMushtak Alright, thank you for your help! Could you answer the question explaining that difference? Perhaps someone else could use that to highlight the difference, if someone else makes the same mistake. –  Feb 07 '16 at 15:26

2 Answers2

2

Let's look at Problem 3.5 again:

3.5 An urn contains 6 white and 9 black balls. If 4 balls are to be randomly selected without replacement, what is the probability that the first 2 selected are white and the last 2 black?

Here, the only thing given to you is the situation that there are $6$ white and $9$ black balls and that $4$ balls are going to be picked randomly. All of this simply tells you the situation of the event, but there is no specific condition for which we can use conditional probability to solve this problem. Thus, when you divide by the probability that the first two selected are white, which is $P(F)=\frac 1 7$, you get an answer that is that is $7$ times the normal answer because the real answer is simply $P(E \cap F)$. It is not given to you that the probability that the first two selected are white is going to happen, so you can not use that as a condition in conditional probability. This problem asks for the probability only given a situation without any special condition, so conditional probability is not applicable.

Now, let's look at Problem 3.3:

3.3 Use Equation (2.1) to compute in a hand of bridge the conditional probability that East has 3 spades given that North and South have a combined total of 8 spades.

Here, the given condition is that "North and South have a combined total of 8 spades." This is not part of the situation, but is rather a condition that we must account for. Thus, when you use conditional probability to account for this condition, you get the right answer because here, conditional probability is applicable.

Noble Mushtak
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Method 1.

$$ \text{Probability}={\frac{{{6}\choose{2}}{{9}\choose{2}}}{{{15}\choose{2}}{{13}\choose{2}}}}=\frac{6}{91}. $$

$$ \binom{6}{2}\text{ is the quantity of ways to choose 2 any white balls of 6 white balls}, $$

$$ \binom{9}{2}\text{ is the quantity of ways to choose 2 any black balls of 9 black balls}, $$

$$ \binom{15}{2}\text{ is the quantity of ways to choose 2 any white balls of} $$

$$ \text{ the total amount of balls (15)}, $$

$$ \binom{13}{2}\text{ is the quantity of ways to choose 2 any black balls of} $$

$$ \text{ the total amount of balls (13), considering that 2 any white balls are chosen already}. $$

Method 2.

Also this problem can be performed using the more detailed binoms:

$$ \text{Probability}=\frac{\binom{6}{1}\binom{5}{1}\binom{9}{1}\binom{8}{1}}{\binom{15}{1}\binom{14}{1}\binom{13}{1}\binom{12}{1}}=\frac{6}{91}. $$

Vadim Chernetsov
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