If I draw 2 balls from the box with 6 white balls and 4 red balls, what is probability that second ball is red?

Question

There is a question bugging me about counting / probability. So the question goes:

If I draw 2 balls from the box with 6 white balls and 4 red balls, what is the probability that 2nd ball is red?

Before asking the problem, I have a side-question. Is this statement same as:

What is the probability that at least one ball is red?

Back to the original question, to solve this, I have this equation. \begin{align} p(R_2) = p(R_2\cap W_1) + p(R_2 \cap R_1) \end{align} For clarification of the notation, $R_1$ means first ball is red $W_2$ mean second ball is white, and so on.

For this, I have two approaches.

1. Using conditional probability: \begin{align} p(R_2) &= p(R_2\cap W_1) &+& p(R_2 \cap R_1)\\ &=p(W_1)p(R_2|W_1) &+& p(R_1)p(R_2|R_1) \\ &=\frac{6}{10}\times \frac{4}{9} &+& \frac{4}{10}\times \frac{3}{9}\\ &=\frac{2}{5} \end{align}

2. Using counting techniques: \begin{align} p(R_2) &= p(R_2\cap W_1) &+& p(R_2 \cap R_1)\\ &= p(W_1\cap R_2) &+& p(R_1 \cap R_2)\\ &= \frac{_6C_1}{_{10}C_2} \times\frac{_4C_1}{_{10}C_2}&+&\frac{_6C_0}{_{10}C_2} \times\frac{_4C_2}{_{10}C_2}\\ &=\frac{24}{45} &+& \frac{6}{45}\\ &=\frac{2}{3} \end{align}

So I think both approaches make sense to me, but the results are different. And I don't know which one is the right one, or neither.

Please help me with this problem and I wonder how I should think through when I have this kind of question.

Answer 1

The two questions you state are different because in the second one you get $R_1,W_2$ into consideration while in the first one, you don't. As for the solutions, probabilistic one is the correct one (i.e. the first one).

Answer 2

Your first answer is correct.

Imagine lining up the four red balls and six white balls in a row without looking at the balls. The question is equivalent to asking whether there is a red ball in the second position. Since the red balls are equally likely to be in any position in the sequence, the probability that a red ball is in the second position is simply $$\frac{4}{4 + 6} = \frac{4}{10} = \frac{2}{5}$$

In your second attempt, you did not take into account the order of selection. Also, the numbers you obtained do not match your calculations.

We correct your attempt.

The probability of choosing two red balls when selecting two of the ten balls is $$\Pr(\text{two red}) = \frac{\dbinom{4}{2}}{\dbinom{10}{2}}$$ The probability of choosing a red ball and a white ball when selecting two of the ten balls is $$\Pr(\text{one red and one white}) = \frac{\dbinom{4}{1}\dbinom{6}{1}}{\dbinom{10}{2}}$$ When one red and one white ball are selected, half the time the red ball will be selected first rather than second. Thus, the probability that a red ball is selected second is $$\Pr(\text{red ball is selected second}) = \frac{\dbinom{4}{2}}{\dbinom{10}{2}} + \frac{1}{2} \cdot \frac{\dbinom{4}{1}\dbinom{6}{1}}{\dbinom{10}{2}} = \frac{2}{5}$$