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?
Best Answer
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.