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 \(\frac{9}{22}\).

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

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

Step4: Calculate the probability of choosing the second green grape

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

Step5: Calculate the probability of both events

Since these are independent - like events (the second - event depends on the first), we use the multiplication rule for probability. The probability that both grapes are green is \(\frac{9}{22}\times\frac{8}{21}=\frac{9\times8}{22\times21}=\frac{72}{462}\approx0.156\) or \(15.6\%\).

Answer:

about \(15.6\%\)