1. **Problem statement:** Find the probability that the third ball drawn is red when balls are drawn without replacement from a bag containing 6 red and 9 green balls. 2. **Understanding the problem:** There are initially 15 balls (6 red, 9 green). Two balls are drawn and set aside, then the third ball is drawn. We want the probability the third ball is red. 3. **Key rule:** Since balls are not replaced, the total number of balls and composition changes after each draw. 4. **Calculate the probability:** The third ball can be red in different scenarios depending on the first two draws. We use the law of total probability: $$P(\text{3rd red}) = P(\text{3rd red} | \text{first 2 red})P(\text{first 2 red}) + P(\text{3rd red} | \text{1 red 1 green})P(\text{1 red 1 green}) + P(\text{3rd red} | \text{2 green})P(\text{2 green})$$ 5. **Calculate each term:** - $P(\text{first 2 red}) = \frac{6}{15} \times \frac{5}{14} = \frac{30}{210} = \frac{1}{7}$ - $P(\text{1 red 1 green}) = \frac{6}{15} \times \frac{9}{14} + \frac{9}{15} \times \frac{6}{14} = \frac{54}{210} + \frac{54}{210} = \frac{108}{210} = \frac{18}{35}$ - $P(\text{2 green}) = \frac{9}{15} \times \frac{8}{14} = \frac{72}{210} = \frac{12}{35}$ 6. **Conditional probabilities for the third ball:** - If first 2 red drawn, remaining red balls = $6 - 2 = 4$, total balls left = $15 - 2 = 13$, so $P(\text{3rd red} | \text{first 2 red}) = \frac{4}{13}$ - If 1 red and 1 green drawn, remaining red balls = $6 - 1 = 5$, total balls left = 13, so $P(\text{3rd red} | \text{1 red 1 green}) = \frac{5}{13}$ - If 2 green drawn, remaining red balls = 6, total balls left = 13, so $P(\text{3rd red} | \text{2 green}) = \frac{6}{13}$ 7. **Calculate total probability:** $$P(\text{3rd red}) = \frac{4}{13} \times \frac{1}{7} + \frac{5}{13} \times \frac{18}{35} + \frac{6}{13} \times \frac{12}{35}$$ 8. **Simplify:** $$= \frac{4}{91} + \frac{90}{455} + \frac{72}{455} = \frac{20}{455} + \frac{90}{455} + \frac{72}{455} = \frac{182}{455}$$ 9. **Final answer:** $$P(\text{3rd red}) = \frac{182}{455} = \frac{26}{65} \approx 0.4$$ --- 1. **Problem statement:** Find the probability that it takes exactly four selections to get four green balls when balls are drawn without replacement from a bag containing 6 red and 9 green balls. 2. **Understanding the problem:** We want the probability that the first time we have drawn 4 green balls is on the 4th draw. This means the first 3 draws contain exactly 3 green balls and 0 red balls (since 4 green balls in 4 draws means all 4 are green). 3. **Key rule:** Since balls are not replaced, probabilities change after each draw. 4. **Calculate the probability:** The event is that the first 4 balls drawn are all green. 5. **Calculate:** $$P(\text{4 green in first 4 draws}) = \frac{9}{15} \times \frac{8}{14} \times \frac{7}{13} \times \frac{6}{12}$$ 6. **Simplify:** $$= \frac{9 \times 8 \times 7 \times 6}{15 \times 14 \times 13 \times 12} = \frac{3024}{32760} = \frac{7}{76} \approx 0.0921$$ --- 1. **Problem statement:** Find the probability that e wears glasses given that they study one language. 2. **Understanding the problem:** This is a conditional probability problem: $P(e \text{ wears glasses} | e \text{ studies one language})$. 3. **Key rule:** Conditional probability formula: $$P(A|B) = \frac{P(A \cap B)}{P(B)}$$ 4. **Given:** - $d$ studies one language and wears glasses (2 marks) - $e$ wears glasses given that they study one language (2 marks) Since no numerical data is provided, we cannot compute exact probabilities. --- **Summary:** - Probability third ball is red: $\frac{26}{65}$ - Probability it takes exactly four selections to get four green balls: $\frac{7}{76}$ - Conditional probability for e wearing glasses given studying one language cannot be computed with given data.