Question

question
a bowl contains 6 green grapes, 10 red grapes, and 8 black grapes. which of the following is the correct
calculation for the probability of choosing a red grape and then without putting the red grape back into
the bowl, choosing a green grape?

○ 10/24 + 6/24
○ 10/24 + 6/23
○ 10/24 × 6/24
○ 10/24 × 6/23

Explanation:

Step1: Calculate total number of grapes initially

Total grapes = \(6 + 10+8=24\). Probability of choosing a red grape first: \(\frac{10}{24}\) (number of red grapes over total grapes).

Step2: Calculate probability of choosing a green grape second

After removing one red grape, total grapes left = \(24 - 1=23\). Probability of choosing a green grape: \(\frac{6}{23}\) (number of green grapes over remaining grapes).

Step3: Use multiplication rule for dependent events

Since these are dependent events (second event depends on the first - no replacement), we multiply the probabilities. So the calculation is \(\frac{10}{24}\times\frac{6}{23}\)

Answer:

\(\frac{10}{24}\times\frac{6}{23}\) (the fourth option)