1. **Problem Statement:** A basket contains 10 ripe and 4 unripe bananas. Four bananas are taken one after another. We want to find the possible values of the random variable $R$ representing the number of ripe bananas selected. 2. **Step 1: List the sample space (S)** The sample space consists of all possible outcomes of selecting 4 bananas from 14 (10 ripe + 4 unripe). 3. **Step 2: Count the number of ripe bananas ($R$) in each outcome** Possible values of $R$ are $0,1,2,3,4$ since we can select from none to all ripe bananas. 4. **Step 3: Frequency description** Calculate the number of ways to select $k$ ripe bananas and $4-k$ unripe bananas: $$\text{Frequency}(R=k) = \binom{10}{k} \times \binom{4}{4-k}$$ for $k=0,1,2,3,4$. 5. **Step 4: Probability description** Total number of ways to select 4 bananas from 14: $$\binom{14}{4}$$ Probability for each $k$: $$P(R=k) = \frac{\binom{10}{k} \times \binom{4}{4-k}}{\binom{14}{4}}$$ 6. **Calculate each probability:** - $P(R=0) = \frac{\binom{10}{0} \binom{4}{4}}{\binom{14}{4}} = \frac{1 \times 1}{1001} = \frac{1}{1001}$ - $P(R=1) = \frac{\binom{10}{1} \binom{4}{3}}{1001} = \frac{10 \times 4}{1001} = \frac{40}{1001}$ - $P(R=2) = \frac{\binom{10}{2} \binom{4}{2}}{1001} = \frac{45 \times 6}{1001} = \frac{270}{1001}$ - $P(R=3) = \frac{\binom{10}{3} \binom{4}{1}}{1001} = \frac{120 \times 4}{1001} = \frac{480}{1001}$ - $P(R=4) = \frac{\binom{10}{4} \binom{4}{0}}{1001} = \frac{210 \times 1}{1001} = \frac{210}{1001}$ 7. **Summary Table:** | Value of $R$ | Frequency | Probability | |--------------|-----------|-------------| | 0 | 1 | $\frac{1}{1001}$ | | 1 | 40 | $\frac{40}{1001}$ | | 2 | 270 | $\frac{270}{1001}$ | | 3 | 480 | $\frac{480}{1001}$ | | 4 | 210 | $\frac{210}{1001}$ | --- **Homework Problem:** - Basket: 10 unripe, 4 ripe bananas - Select 3 bananas - Find possible values of $U$ = number of unripe bananas selected This problem is similar; possible values of $U$ are $0,1,2,3$. Frequency: $$\text{Frequency}(U=k) = \binom{10}{k} \times \binom{4}{3-k}$$ Probability: $$P(U=k) = \frac{\binom{10}{k} \times \binom{4}{3-k}}{\binom{14}{3}}$$ This completes the solution for the first problem.