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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?

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3 Answers

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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:

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

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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 $\displaystyle \frac{21}{45} = \frac{7}{15}$.

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

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