Question 1083927: From a box containing 25 item 5 of which are defective, 4 are chosern at random ,let X be the number of defective items found.Obtain the probability distribution of X if
I. the items are chosen with replacement.
II. the items are chosen with out replacement.
Found 2 solutions by Boreal, natolino_2017:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! With replacement, the probability of
0 defective is (4/5)^5=0.328
1 defective is 5(4/5)^4(1/5)
2 defective is 10(4/5)^3(1/5)^2
3 defective is 10(4/5)^2(1/5)^3
4 defective is 5(4/5)(1/5)^4
5 defective is (1/5)^5
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without replacement
0 defective is (20/25)(19)24)(18/23)(17/22)(16/21) or 20C5/25C5=20!/15!5! divided by 25!/20!5!=0.2918
1 defective is 20C4*5C1/25C5
2 defective is 20C3*5C2/25C5
3 defective is 20C2*5C3/25C5
4 defective is 20C1*5C4/25C5
5 defective is 1/25C5=5/25*4/24*3/23*2/22*1/21=5C5/25C5=1/25!/5!20!=5!20!/25!=the first part.
Answer by natolino_2017(77)  (Show Source):
You can put this solution on YOUR website! I) with replacement
x: number of defective in 4 picking.
Range of x ={0,1,2,3,4}
Every picking is independent from the rest.
p= P(pick one defective) = number of defective/total items = 5/25 =1/5.
x is binomial (n= 4, p=1/5).
P(x=0) = 4C0(p^0)(1-p)^(4-0) = 256/625.
P(x=1) = 4C1(p^1)(1-p)^(4-1) = 256/625.
P(x=2) = 4C2(p^2)(1-p)^(4-2) = 96/625.
P(x=3) = 4C3(p^3)(1-p)^(4-3) = 16/625.
P(x=4) = 4C4(p^4)(1-p)^(4-4) = 1/625.
II) without replacement
y: number of defective in 4 picking.
Range of y ={0,1,2,3,4}
y is hyper-geometric with N=25, n=4, d=5.
P(y=0) = (5C0)((25-5)C(4-0))/(25C4) = 969/2530.
P(y=1) = (5C1)((25-5)C(4-1))/(25C4) = 114/253
P(y=2) = (5C2)((25-5)C(4-2))/(25C4) = 38/253
P(y=3) = (5C3)((25-5)C(4-3))/(25C4) = 4/253
P(y=4) = (5C4)((25-5)C(4-4))/(25C4) = 1/2530
check for yourself that, in both results, the sum of every probabilities is equal to one.
@natolino_
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