p good = .2
p not good = .8
this is a binomial distribution problem
P(n,k) = C(n,k) p^k (1-p)^(n-k)
where
C(n,k) = n!/[k!(n-k)!]
here for example
P(10,1) = C(10,1).2^1 .8^9
C(10,1)= 10!/[1(9)!] = 10
so
P(10,1) = 10 * .2 * .8^9
= .268
similarly
C(10,2) = 10!/[2(8!] = 10*9/2 =45
so
P(10,2) = 45 .2^2 .8^8
= .302
A factory finds out that on the average 20% of bolts produced by the given machine will be defective.if 10 bolts are selected at random from the day's production of this machine,find the probability that 1.exactly 2 will be defective.
(this question is base on probability and combinatorial Analysis)
1 answer
1 answer