1. **Problem Statement:** In a group of five potential blood donors—a, b, c, d, and e—only a and b have type O-positive blood. Five blood samples, one from each individual, will be typed in random order until an O+ individual is identified. Let the random variable $Y$ = the number of typings necessary to identify an O+ individual. Determine the probability mass function (pmf) of $Y$. 2. **Understanding the problem:** We have 5 individuals: a, b (O+), c, d, e (not O+). We randomly order the 5 samples and test them one by one until we find an O+ individual. $Y$ is the position in the sequence where the first O+ is found. 3. **Key points:** - There are 2 O+ individuals out of 5. - The order is random and all permutations are equally likely. - $Y$ can take values 1 through 4 because the first O+ must appear by the 4th test at the latest (since there are only 2 O+ individuals). 4. **Formula for pmf:** The pmf for the position of the first success in sampling without replacement from a finite population with $k$ successes and $N-k$ failures is given by the negative hypergeometric distribution: $$P(Y = y) = \frac{\binom{y-1}{0} \binom{N-y}{k-1}}{\binom{N}{k}}$$ where: - $N=5$ total individuals - $k=2$ successes (O+) - $y$ is the number of trials until the first success 5. **Calculate pmf values:** - For $y=1$: $$P(Y=1) = \frac{\binom{0}{0} \binom{4}{1}}{\binom{5}{2}} = \frac{1 \times 4}{10} = 0.4$$ - For $y=2$: $$P(Y=2) = \frac{\binom{1}{0} \binom{3}{1}}{10} = \frac{1 \times 3}{10} = 0.3$$ - For $y=3$: $$P(Y=3) = \frac{\binom{2}{0} \binom{2}{1}}{10} = \frac{1 \times 2}{10} = 0.2$$ - For $y=4$: $$P(Y=4) = \frac{\binom{3}{0} \binom{1}{1}}{10} = \frac{1 \times 1}{10} = 0.1$$ - For $y=5$: Since there are only 2 O+ individuals, the first O+ must appear by the 4th test, so $P(Y=5) = 0$. 6. **Check sum:** $$0.4 + 0.3 + 0.2 + 0.1 = 1$$ 7. **Interpretation:** - There is a 40% chance the first tested individual is O+. - 30% chance the first O+ is found on the second test, and so on. **Final pmf:** $$P(Y=y) = \begin{cases} 0.4 & y=1 \\ 0.3 & y=2 \\ 0.2 & y=3 \\ 0.1 & y=4 \\ 0 & \text{otherwise} \end{cases}$$