himanshuhpr wrote:
D = defective, F = functional
I'm having trouble in comprehending why is it in 4 different ways, why is the order important as we are just selecting randomly 4 cars ... so DFFF, FDFF, FFDF, FFFD should accordingly be just 1 case .. ???
In a shipment of 20 cars, 3 are found to be defective. If four cars are selected at random, what is the probability that exactly one of the four will be defective?(A) 170/1615
(B) 3/20
(C) 8/19
(D) 3/5
(E) 4/5
APPROACH #1:\(P=\frac{C^1_3*C^3_{17}}{C^4_{20}}=\frac{8}{19}\), where \(C^1_3\) is number of ways to select 1 defective car out of 3, \(C^3_{17}\) is number of ways to select 3 not defective car out of 17, and \(C^4_{20}\) is total number of ways to select 4 cars out of 20.
Answer: C.
APPROACH #2:We need the probability of DFFF: \(P(DFFF)=\frac{4!}{3!}*\frac{3}{20}*\frac{17}{19}*\frac{16}{18}*\frac{15}{17}\). We are multiplying by \(\frac{4!}{3!}=4\), since DFFF case can occur in 4 ways: DFFF (the first car is defective and the second, third and fourth are not), FDFF, FFDF, FFFD (basically number of permutations of 4 letter DFFF out of which 3 N's are identical).
Answer: C.
Hope it's clear.
I have a doubt here: the question assumes that you select the cars sequentially, i.e. one after the other. But what if I decided to select the 4 cars at the same time? Could you use the binomial distribution to model the problem?