a bowl has 9 green grapes and 13 red grapes. henry randomly chooses a grape, eats it, and then chooses…

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

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

Answer

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%)