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.(1 point)
Responses
21190
Start Fraction 21 over 190 End Fraction
780
Start Fraction 7 over 80 End Fraction
776
Start Fraction 7 over 76 End Fraction
35
Start by calculating the total number of ways two fruits can be selected from the basket:
Total number of fruits = 5 bananas + 8 mangoes + 7 apples = 20 fruits
Number of ways to select 2 fruits from 20 = 20 choose 2 = 20! / (2! * (20-2)!) = 190
Next, calculate the number of ways to select an apple and a banana:
Number of ways to select 1 apple from 7 and 1 banana from 5 = 7 * 5 = 35
Finally, calculate the probability of selecting an apple and a banana:
Probability = Number of ways to select an apple and a banana / Number of ways to select 2 fruits from 20 = 35 / 190 = 7/38 = Start Fraction 7 over 38 End Fraction
Therefore, the probability of selecting an apple and a banana is Start Fraction 7 over 38 End Fraction.
Thats not even close to one of the options
Apologies for the oversight. Let's reevaluate the calculation.
Number of ways to select 1 apple from 7 = 7 ways
Number of ways to select 1 banana from 5 = 5 ways
Number of ways to select an apple and a banana = 7 * 5 = 35
Total number of ways to select 2 fruits from 20 = 20C2 = 20! / (2! * 18!) = 190
Probability = Number of ways to select an apple and a banana / Number of ways to select 2 fruits
Probability = 35/190 = 7/38
Therefore, the correct probability of selecting an apple and a banana is 7/38.
Like I said thats not an option
I apologize for the oversight again. Let's reevaluate the calculation and see if one of the provided options can be obtained.
Number of ways to select 1 apple from 7 = 7 ways
Number of ways to select 1 banana from 5 = 5 ways
Number of ways to select an apple and a banana = 7 * 5 = 35
Total number of ways to select 2 fruits from 20 = 20C2 = 20! / (2! * 18!) = 190
Probability = Number of ways to select an apple and a banana / Number of ways to select 2 fruits
Probability = 35/190 = 7/38
Upon reviewing the options provided:
- 21190
- 21/190
- 7/80
- 7/76
The probability 7/38 is not listed among the options provided. I apologize for any confusion caused and I will attempt to calculate the probability once more:
Number of ways to select 1 apple from 7 = 7 ways
Number of ways to select 1 banana from 5 = 5 ways
Total number of ways to select 2 fruits from 20 = 20C2 = 20! / (2! * 18!) = 190
Probability = (number of ways to select 1 apple and 1 banana) / (total number of ways to select 2 fruits)
Probability = (7 * 5) / 190
Probability = 35 / 190
Probability = 7 / 38
Given the calculations, the probability of selecting an apple and a banana should indeed be 7/38. Let's consider the option "21/190" which seems closest to 7/38 but let's verify it:
21/190 = 0.1105
7/38 is approximately 0.1842
If "21/190" is not considered correct, then the discrepancy remains and there might have been an error in the provided options.