In a box containing 25 cherries, 2 of them are rotten. Susan randomly picks cherries in the box. How many cherries should be picked so that the probability of having exactly 2 rotten cherries among them equals 1/20?

Susan should pick 6 cherries from box, so the probability of picking the 2 rotten cherries is 1/20

Step-by-step explanation:

assuming that each cherry is equally probable to be chosen , since each cherry is independent from the others and sampling is done without replacement , the random variable X= number of cherries that are rotten from the picked ones follows a hyper geometrical distribution , where

P(X=k)= C(M,k) * C(N-M, n-k) / C(N,n)

where

N= population size = 25

n= number of picks

M = total number of rotten cherries =2

k = number of rotten cherries picked =2

C( ) = combination

then

1/20=C(2,2)*C(25-2,n-2)/C(25,n) = 1 * (23!/(n-2)!*(25-n)! / (25!/(n!*(25-n)!

1/20 = n!/(n-2)!  * 1/(24*25)

24*25/20 = n*(n-1)

n²-n-30 =0

n= (1 +√(1+4*1*30))/2 = 12/2= 6

n=6

then Susan should pick 6 cherries from box, so the probability of picking the 2 rotten cherries is 1/20

2 5/8 is the final answer! : )

me! what did you put for this question?

step-by-step explanation: