Question

A sample of 4 different calculators is randomly selected from a group containing 18 that are defective and 35 that have no defects. what is the probability that at least one of the calculators is defective?

Answer 1

1.

P(at least one of the calculators is defective)= 

1- P(none of the selected calculators is defective).

2.

P(none of the selected calculators is defective)

           =n(ways of selecting 4 non-defective calculators)/n(total selections of 4)

3.

selecting 4 non-defective calculators can be done in C(35, 4) many ways, 

where  $C(35, 4)= \frac{35!}{4!31!}= \frac{35*34*33*32*31!}{4!*31!}= \frac{35*34*33*32}{4!}= \frac{35*34*33*32}{4*3*2*1}= 52,360$

while, the total number selections of 4 out of 18+35=53 calculators can be done in C(53, 4) many ways,

$C(53, 4)= \frac{53!}{4!*49!}= \frac{53*52*51*50}{4*3*2*1}= 292,825$

4. so, P(none of the selected calculators is defective)=$\frac{52,360}{292,825} =0.18$


5. P(at least one of the calculators is defective)= 

1- P(none of the selected calculators is defective)=1-0.18=0.82



Answer:0.82