First, calculate the total number of ways to select 2 fruits from a basket of 20 fruits (5 bananas + 8 mangoes + 7 apples).
There are a total of C(20, 2) = 20! / (2! * 18!) = 190 ways to select 2 fruits from the basket.
Next, calculate the probability of selecting an apple and a banana.
There are a total of 5 bananas and 7 apples, so the number of ways to select an apple and a banana is C(5, 1) * C(7, 1) = 5 * 7 = 35 ways.
Therefore, the probability of selecting an apple and a banana is 35 / 190 = 7 / 38.
The correct answer is not provided in the options.
A basket of fruits contains 5 bananas, 8 mangoes, and 7 apples. Two fruits are randomly selected without replacement. Calculate the probability of selecting an apple and a banana.
A. 3/5
B. 21/190
C. 7/76
D. 7/80
3 answers
yes it is.
Apologies for the oversight. Let's reevaluate the probability calculation:
The total number of ways to select 2 fruits from the basket of 20 fruits is C(20, 2) = 190.
The number of ways to select 1 apple from the 7 available apples is C(7, 1) = 7.
The number of ways to select 1 banana from the 5 available bananas is C(5, 1) = 5.
Therefore, the number of ways to select 1 apple and 1 banana is 7 * 5 = 35.
The probability of selecting an apple and a banana is 35 / 190 = 7 / 38.
Therefore, the correct answer is not among the options provided. Apologies for the confusion.
The total number of ways to select 2 fruits from the basket of 20 fruits is C(20, 2) = 190.
The number of ways to select 1 apple from the 7 available apples is C(7, 1) = 7.
The number of ways to select 1 banana from the 5 available bananas is C(5, 1) = 5.
Therefore, the number of ways to select 1 apple and 1 banana is 7 * 5 = 35.
The probability of selecting an apple and a banana is 35 / 190 = 7 / 38.
Therefore, the correct answer is not among the options provided. Apologies for the confusion.