In a shipment of 20 cars, 3 are found to be defective. If

Question

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

Bunuel · Accepted Answer

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.

anindyat · Answer

Out of 3 defective cars â€“ select 1 => 3 possible ways
Out of 17 good cars â€“ select 3 => 17 C 3 possible ways

Probability = (3 * 17 C 3) / (20 C 4) = 8 /19

kidderek · Answer

Different yet very effective. My approach is more convoluted

I did 4C1 (3/20 * 17/19 * 16/18 * 15/17) = 48,960/116,280

I was reducing the fraction and got stuck at 136/323. Couldn't reduce it for the life of me. 8/19 looked closest so I divided the numerator and denominator by 8/19 and got 17 as the divisor.

chile · Answer

D : Defective car
G : OK car

The posibilities are:

GGGD
GGDG
GDGG
DGGG

The probability for each possibility is:

17/20*16/19*15/18*3/17
17/20*16/19*3/18*15/17
17/20*3/19*16/18*15/17
3/20*17/19*16/18*15/17

The answer for this problem is the sum of the probabilities of all possibilities

4*(17*16*15*3)/(20*19*18*17) = 8/19

ywilfred · Answer

D - defective
N - non-defective

DNNN --> P = (3/20)(17/19)(16/18)(15/17) = 2/19
NDNN --> P = (17/20)(3/19)(16/18)(15/17) = 2/19
NNDN --> 2/19
NNND --> 2/19

Total = 4 * 2/19 = 8/19

Ans C

Damager · Answer

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

chances of 1st to be defective and rest not def
(3/20)*(17/19)*(16/18)*(15/17)

Prob for 2nd def rest ok is same ..

so 4*((3/20)*(17/19)*(16/18)*(15/17))= 8/19

Juaz · Answer

4*(3/20*17/19*16/18*15/17)
4*(3/18*16/20*15/19)
4*(4/6*3/19)
4*(2/19) (its easy to do on paper as you can mark out).

amorica · Answer

I'm having trouble w/ the numerator on this one..

I know that the denominator is 20c4.... 

but what are the numbe of combinations of having one defect?

GGGD
GGDG
GDGG
DGGG

wouldn't each defect (out of the 3) have the same grouping above? So the total would be 12? 

Thanks for anyone who can set me straight.

AK · Answer

i think u r confusing with permutation .... the sequence of defect is not important here.

numerator = ( 1 out of three defective) and ( 3 out of 17 good ones)
 = 3C1 * 17C3

himanshuhpr · Answer

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 .. ???

louis3092 · Answer

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?

Bunuel · Answer

Mathematically the probability of picking simultaneously, or picking them one at a time (without replacement) is the same.

Archit3110 · Answer

defective cars = 3 and not defective = 17
for 4 cars chosen at random
3/20 * 17/19*16/18 * 15/17 = 2/19
and these 4 cars can be arranged in 4!/3! ways ; 4 ways
4*2/19 ; 8/19
OPTION C

RastogiSarthak99 · Answer

P(d) = 3/20; p(nd) = 17/20

Keeping exactly 1 so, choosing 4 would mean

3/20 * 17/19 * 16/18 * 15/17 = 2/19

2/19 * No of arrangements

2/19 * 4!/3! * 1!

=8/19

or 2/19 + 2/19 + 2/19 + 2/19 = 8/19

darunp · Answer

Think of probability as probability = no. of desire outcomes/no. of total outcomes

total outcomes: choosing 4 out of 20 cars
=  \frac{20!}{16!*4!}= \frac{20*19*18*17}{4*3*2*1} = 19*17*5*3

desired outcome: exactly 1 of 4 is defective

choose 1 defective out of 3 =  \frac{3!}{2!*1!} = 3

choose 3 cars out of remaining 17 cars =  \frac{17!}{14!*3!} = \frac{17*16*15}{3*2*1} = 17*8*5

total ways to choose 1 defective out of 4 =  17*8*5*3

probability = \frac{17*8*5*3}{19*17*5*3} = \frac{8}{9}

CapnSal · Answer

Can this be solved via the binomial formula of (nPr).P^r(1-p)^n-r? Bunuel

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