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I have an urn with 10 balls: 4 red and 6 white. What is the probability that the first two drawn balls have the same color?

Approach No. 1:

$|U| = 10! = 3628800$

White balls on first and second draw: on the first and second draw, I can pick any of the 6 white balls on the first draw, and 5 on the second draw. After that, I have 8 balls left, of which I don't care what order they are in. This will become $6\times5\times8\times7\times6\times5\times4\times3\times2\times1=1209600$.

$\left(\frac{1209600}{3628800} \right)=\frac{1}{3}$

If we have red balls on the first and second draw, then: I can pick any of the 4 red balls on the first draw, and on the second draw any of the left over 3. This will be$4\times3\times8\times7\times6\times5\times4\times3\times2\times1=483840$.

$\left(\frac{483840}{3628800} \right)=\frac{2}{15}$

The sum of these two will be: $\frac{1}{3}+\frac{2}{15}=\frac{7}{15}$

Approach number two:

Chance of a red ball on first draw: $\frac{4}{10}$. Chances of a red ball on second draw: $\frac{3}{9}.$

$\frac{4}{10}\times\frac{3}{9}=\frac{2}{15}$.

Chance of a white ball on first draw: $\frac{6}{10}$. Chances of a white ball on second draw: $\frac{5}{9}$.

$\frac{6}{10}\times\frac{5}{9}=\frac{5}{15}$.

$\frac{5}{15}+\frac{2}{15}=\frac{7}{15}$

What I want to know, is this answer correct? Somehow I can't get it straight in my head that this is the right answer, since the $8!$ seems a bit weird to me. I mean don't the chances of the other red balls in the case of red balls on first and second draw count?

rschwieb
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Garth Marenghi
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3 Answers3

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Yes this is correct.

With conditional probability, you could formulate it this way: P(2 balls of the same color) = P(2 balls are white OR 2 balls are red) = P(2 balls are white) + P (2 balls are red) [since the two conditions are mutually exclusive]

P(2 balls are white) = P(first ball is white AND (second ball is white given first ball is white)) = P(first ball is white) * P(second ball is white | first ball is white) = 6/10 * 5/9, as you mentioned.

In the same way, P(2 balls are red) = 4/10 * 3/9.

The two rules that we use are described here: http://people.richland.edu/james/lecture/m170/ch05-rul.html

It is 1) the addition of mutually exclusive events and 2) dependent events and contitional probabilities.

mika
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  • Right, this is very clear, thank you! I think the confusion for me is that the question asks for "calculate the chances that the first two drawn balls are the same color". What they probably mean is: the chances you draw a ball of a certain color for the very first time, and the second time you draw one of the same color. E.g., for the chances of two white balls, it could be: rrWrWrwrwwww (where caps W are the ones we will calculate the chances for), while I was thinking it had to start with two white balls. So when you draw the very first time, it had to be white, and the second one as well. – Garth Marenghi Aug 21 '13 at 14:39
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As Doctor Dan says, the second approach is also correct and perhaps more intuitive.

A third approach might be to consider that sampling without replacement gives you a randomly chosen subset of two balls drawn from the $10$ balls in the urn. So we have $\dbinom{10}{2} = 45$ possible outcomes of the experiment of which $\dbinom{4}{2} = 6$ subsets have two red balls and $\dbinom{6}{2} = 15$ subsets have two white balls so that the probability of getting two balls of the same color is $\displaystyle \frac{6+15}{45} =\frac{21}{45}= \frac{7}{15}$ as you have found already.

It is also useful to check the calculations by noting that there must be $4\times 6 = 24$ subsets with balls of different colors since there are $4$ choices of red ball and $6$ of white ball to include in the set, and $24 = 45-21$. Some might even say that this last way is an even easier way of getting at the desired answer of $\displaystyle \frac{21}{45} = \frac{7}{15}$.

Dilip Sarwate
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  • That's a nice way of doing this as well. As I said to mika, my confusion about the correctness of the answer comes from how the question is asked: "calculate the chances that the first two drawn balls are the same color". But I think that it doesn't necessarily need to be on the very first or second draw, or even consecutive, just the first and second time that you draw the balls of the same color. – Garth Marenghi Aug 21 '13 at 14:54
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S = {Set of all possible combination of 2 balls from the available 4 Red and 6 White Balls}

n(S) = $^{10} P_{2}$ ways = $10 \times 9 = 90$ (Since, 2 balls can be selected from $10$ in $^{10} P_{2}$ ways)

Now,

E = Both are of same colour

n(E) = $2$ Red balls can be selected from $4$ in $^4 P_{2}$ ways AND $2$ White balls can be selected from $6$ in $^6 P_{2}$ ways) Total = $^4 P_{2} + ^6 P_{2} =(12 + 30) =42$ ways.

P(E) = $\frac{n(E)}{n(S)} = \frac{42}{90} = \frac{7}{15}$.

Jiggu
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