Answer:
0.041
Step-by-step explanation:
Number of rotten apples = 4
Total number of apples= 15
Number of good apples = 15-4
= 11
If the 9th apple is the last rotten apple, the first 8 apples picked should have 3 rotten apples and 5 good apples. This selection can be done by
C(4,3) * C(11,5)
Recall that C(n,r) = n! /(n-r)! r!
C(4,3) = 4!/(4-3)!3!
= 4!/1!3!
= 4
C(11,5)= 11!/(11-5)!5!
= 11!/ 6!5!
= 462
To select 8 apples from 15 apples, we have C(15,8)
= 15!/(15-8)!8!
= 15!/7!8!
= 6435
The 9th apple which is the rotten apple remaining can be selected in one way out of (15-8) = 7apples
C(7,1) = 7!/(7-1)!1!
= 7!/6!1!
= 7
Total probability = [C(4,3) * C(11,5)] / [C(15,8) * C(7,1)]
= (4*462)/(6435*7)
= 1848/45045
= 0.041
