A bag contains four red counters and six black counters. A counter is picked at random and not replaced. A second counter is then picked. Find the probability that the second counter is red, given that the first counter is red.

Question

anonymous · Answer

3/9  If you have a red counter, there are 3 out of nine red counters left.

anonymous · Answer

Sir, I am trying to do this using the conditional probability rule: P(AnB)/P(B). But I am stuck. Does conditional probability even apply here? If ne, why doesn't it?

kropot72 · Answer

The hypergeometric distribution applies.
$$\frac{\left(\begin{matrix}4 \ 2\end{matrix}\right)\left(\begin{matrix}6 \ 0\end{matrix}\right)}{\left(\begin{matrix}10 \ 2\end{matrix}\right)}$$
$$=\frac{4\times 3\times 2}{2\times 10 \times9}$$

directrix · Answer

attachment

directrix · Answer

P (A and B ) = (4/10 * 3/9)
P(A) = (4/10)
------------
P(B/A) = P ( A and B) / P ( A)
P(B/A) = (4/10 * 3/9) / (4/10) = (4/10 * 3/9) * 10/4 = 3/9 = 1/3

@Shahan03

kropot72 · Answer

@directrix This problem is based on sampling without replacement. It is not a conditional probability problem.