There is a box containing 6 red balls and 5 white balls. If the first drawn ball is red and not returned to the box, find the probability that a white ball is taken in the second drawn.

I have 3 solutions.The first two solutions produce the same but not for the last one. What is wrong with my reasoning?

## Solution 1 (simple logic)

After the first drawn there are 5 red balls and 5 white balls. The probability to take a white ball is $5/10 = 1/2$.

## Solution 2 (conditional probability)

Let \begin{align*} A&=\{(x,y)\mid x \text{ is a red ball and } y \text{ is any ball} \}\\ A\cap B &=\{(x,y)\mid x \text{ is a red ball and } y \text{ is white ball} \}\\ S&=\{(x,y)\mid x \text{ is any ball and } y \text{ is any ball} \} \end{align*}

\begin{align*} n(A)&=6\times 10\\ n(A\cap B)&=6\times 5\\ n(S)&=11\times 10 \end{align*}

\begin{align*} P(A)&=\frac{6\times 10}{11\times 10}\\ P(A\cap B)&=\frac{6\times 5}{11\times 10} \end{align*}

\begin{align*} P(B\mid A)&=\frac{P(A\cap B)}{P(A)}\\ &= \frac{\frac{6\times 5}{11\times 10}}{\frac{6\times 10}{11\times 10}}\\ &=\frac12 \end{align*}

## Solution 3 (permutation point of view)

All balls have numbers to uniquely identify each other. The first and second drawn balls are placed separately in a different place each.

The number of elements in the sample space is $11\times 10$ (there are 11 ways to occupy the first place and 10 ways to occupy the second place). The number elements of the space of event in question is $6\times 5$ (there are 6 ways for red balls to occupy the first place and 5 ways for the white balls to the second place). The probability is $\frac{6\times 5}{11\times 10} = \frac{3}{11}$.

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I don't see why you have 11 choices for the first place; you know it is a red ball, so it can only be one of 6. You're not just choosing any of 11 balls; you know it is red. – user99680 Nov 5 '13 at 4:00

In your solution $3$ the sample space conists of only $6 \times 10$ possibilities, as the first ball must be red.