Question

a bowl has 9 green grapes and 13 red grapes. henry randomly chooses a grape, eats it, and then chooses another grape.
what is the probability that both grapes are green?
about 17.5%
about 16.7%
about 15.6%
about 14.9%

Explanation:

Step1: Calculate the total number of grapes initially

The total number of grapes initially is \(9 + 13=22\).

Step2: Calculate the probability of choosing the first green grape

The probability of choosing the first green grape is \(P_1=\frac{9}{22}\).

Step3: Calculate the number of grapes and green grapes after eating the first grape

After eating one green grape, the number of grapes left is \(22 - 1 = 21\), and the number of green grapes left is \(9- 1=8\).

Step4: Calculate the probability of choosing the second green grape

The probability of choosing the second green grape is \(P_2=\frac{8}{21}\).

Step5: Calculate the probability of both events (using the multiplication rule for dependent events \(P = P_1\times P_2\))

\(P=\frac{9}{22}\times\frac{8}{21}=\frac{9\times8}{22\times21}=\frac{72}{462}\approx0.156\) or \(15.6\%\)

Answer:

about \(15.6\%\)