Question 252819
Please help in answering this question. 
The quality assurance engineer of a television manufacturer inspects TVs in lots of 100. He selects 5 of the 100 TVs at random and inspects them thoroughly.
----
Binomial: n = 5 ; p = 0.06 ; x varies
------
 Assuming that 6 of the 100 TVs the current lot are defective, find the probability that the number of defective TVs obtained by the engineer is
a. exactly one
P(x=1) = 5C1(0.06)(0.94)^4 = 0.2342
--------------------------------
b. at most one
P(x=0 or 1) = 0.7339 + 0.2342 = 0.9681
----------------------------------
c. at least one
P(x>=1) = 1- P(x=0) 
1 - 5C0(0.06)^0*(0.94)^5 
= 1 - 1*1*(0.94)^5 
= 1 - 0.7339 
= 0.2661
----------------------------------
Cheers,
Stan H.