Jelly Bean Odds 80Fb87

1. **State the problem:** We have $6$ jelly beans, $1$ of which is red. We pull out $2$ jelly beans with replacement. We want to find the odds of pulling at least one red jelly bean. 2. **Understand the problem:** "With replacement" means after pulling a jelly bean, we put it back before pulling the next one. So the probability of each pull remains the same. 3. **Formula used:** The probability of at least one red in two pulls is $$P(\text{at least one red}) = 1 - P(\text{no red in both pulls})$$ 4. **Calculate probability of no red in one pull:** Since there is $1$ red and $5$ non-red jelly beans out of $6$, $$P(\text{no red in one pull}) = \frac{5}{6}$$ 5. **Calculate probability of no red in both pulls (with replacement):** $$P(\text{no red in both pulls}) = \left(\frac{5}{6}\right)^2 = \frac{25}{36}$$ 6. **Calculate probability of at least one red:** $$P(\text{at least one red}) = 1 - \frac{25}{36} = \frac{36}{36} - \frac{25}{36} = \frac{11}{36}$$ 7. **Convert probability to odds:** Odds in favor of an event are given by $$\text{Odds} = \frac{P(\text{event})}{1 - P(\text{event})}$$ 8. **Calculate odds of pulling at least one red:** $$\text{Odds} = \frac{\frac{11}{36}}{1 - \frac{11}{36}} = \frac{\frac{11}{36}}{\frac{25}{36}} = \frac{11}{\cancel{36}} \times \frac{\cancel{36}}{25} = \frac{11}{25}$$ **Final answer:** The odds of pulling at least one red jelly bean in two pulls with replacement is $\frac{11}{25}$.