# Urn probability replacement problem

An urn contains $10$ red and $10$ white balls. They are taken out at random one at a time. Find the probability that the fourth white ball is the fourth, fifth, sixth or seventh ball drawn if the sampling is done

A. with replacement
B. without replacement

For A I got that it is $(\frac{1}{2})^4$ but I don't see how the rest of the spaces could be done.

For B I got for the first part $.043$ but once again for the fifth position I try using the conditional probability $\frac{10C_1 \times 10C_3}{20C_4}$ but my answer is not the same answer as the solution.

Any help would be appreciated.

(With replacement) The probability that the fourth white ball is the fourth one drawn, is, as you know, $\frac{1}{2^4}$.
To find the probability it is the fifth ball drawn, note that we must have exactly $3$ white in the first four, and then a white. The probability of this is $\binom{4}{3}\cdot\frac{1}{2^4}\cdot\frac{1}{2}$.
To find the probability it is the sixth ball drawn, note that we must have $3$ white in the first five, and then a white. The probability of this is $\binom{5}{3}\cdot\frac{1}{2^5}\cdot\frac{1}{2}$.
(Without replacement) The basic strategy is the same, but the expressions for the individual probabilities are different. For example, let us compute the probability the fourth white is the sixth ball drawn. We must have $3$ white and $2$ black in the first five trials, and then a white.
The probability of $3$ white and $2$ black in the first five is $\frac{\binom{10}{3}\binom{10}{2}}{\binom{20}{5}}$. Given this has happened, there are $7$ white and $8$ black left, so the probability of a white on the sixth trial is $\frac{7}{15}$. Thus the probability the number of trials is exactly $6$ is $\frac{\binom{10}{3}\binom{10}{2}}{\binom{20}{5}}\cdot\frac{7}{15}$.