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?

Answer 1

Using the it's concept, 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 is:

$p = \frac{5}{2} \times \frac{1}{23} = \frac{5}{46}$

What is a probability?

A probability is given by the number of desired outcomes divided by the number of total outcomes.

In this problem, we have that there is a total of 6 + 10 + 8 = 24 grapes. For the first grape, 10 are red, hence the probability that the first grape is red is:

$pR = \frac{10}{24} = \frac{5}[12}$

For the second grape, considering that the first was red and was removed, there will be 23 grapes, out of which 6 will be green, hence the probability that the second grape is green is:

$pG = \frac{6}{23}$

Hence the correct expression is found as follows:

$p = pR \times pG = \frac{5}{12} \times \frac{6}{23} = \frac{5}{2} \times \frac{1}{23} = \frac{5}{46}$

More can be learned about probabilities at https://brainly.com/question/14398287

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