Henry has a bag that contains strawberry juice, cherry juice and watermelon juice. He performs an experiment. Henry randomly removes a chew from the bag, record the results and return the chew bag perform the experiment 55 times the results are shown below a strawberry was selected 15 times a cherry she was selected 20 times a watermelon. She was selected 20 times based on these results express the probability that the next two he removes from the bag will be cherry or watermelon as a fraction in simplest Form.

1 answer

To find the probability that the next two chews removed from the bag will be cherry or watermelon, we first need to determine the total number of chews and the number of cherry and watermelon chews based on the results of Henry's experiment.

From the experiment:

  • Strawberry chews selected: 15
  • Cherry chews selected: 20
  • Watermelon chews selected: 20

Total = 15 (strawberry) + 20 (cherry) + 20 (watermelon) = 55

Now, we need to find the total number of chews that are either cherry or watermelon:

  • Cherry + Watermelon = 20 + 20 = 40

Next, we can find the probability \( P \) of selecting a cherry or watermelon chew on a single draw: \[ P(\text{Cherry or Watermelon}) = \frac{\text{Number of Cherry or Watermelon}}{\text{Total Number of Chews}} = \frac{40}{55} \]

Now, simplify \( \frac{40}{55} \):

  • Both 40 and 55 can be divided by 5: \[ \frac{40 \div 5}{55 \div 5} = \frac{8}{11} \]

Now we need to find the probability that both chews removed are either cherry or watermelon. Assuming independence between the selections (since the chew is returned to the bag after each draw), the probability that both selected chews are cherry or watermelon is: \[ P(\text{both are Cherry or Watermelon}) = P(\text{Cherry or Watermelon}) \times P(\text{Cherry or Watermelon}) = \left(\frac{8}{11}\right) \times \left(\frac{8}{11}\right) \] \[ = \frac{8 \times 8}{11 \times 11} = \frac{64}{121} \]

Thus, the probability that the next two chews removed from the bag will be cherry or watermelon is: \[ \boxed{\frac{64}{121}} \]