Question 1011187
 
Question:
A shipment of 120 burglar alarms contains 5 that are
defective. If 3 of the alarms are randomly selected and
shipped to a customer, find the probability that the
customer will get one bad unit by using
(i) The formula for the hyper geometric distribution
(ii) The formula for the binomial distribution as an
approximate 
 
Solution:
(i) using hypergeometric distribution
{{{P(x)=C(A,x)*C(B,n-x)/C(A+B,n)}}}
where
x=number of success (defective units)=1
n=size of sample=3
A=number of defective units in batch=5
B=number of non-defective units in batch=120-5=115
C(n,r)=probability of choosing r objects out of n.
{{{P(X=1)=C(5,1)*C(115,2)/C(120,3)}}}
={{{5*6555/280840=6555/56168=0.1167}}} approximately

(ii) using binomial distribution to approximate probability
Here, we assume that the probability of drawing a defective part remains constant at 5/120 over the three draws.
then 
p=5/120=1/24
n=3 (sample size)
x=1 (# of defects)
{{{P(X=1)=C(n,x)*p^(x)*(1-p)^(n-x)}}}
={{{C(3,1)*(1/24)^(1)*(23/24)^(2)}}}
={{{529/4608=0.1148}}} approximately