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)= (35!)/(4!31!)= (35*34*33*32*31!)/(4!*31!)= (35*34*33*32)/(4!)= (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)= (53!)/(4!*49!)= (53*52*51*50)/(4*3*2*1)= 292,825`

4. so, P(none of the selected calculators is defective)=
`(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