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 [tex]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[/tex]
while, the total number selections of 4 out of 18+35=53 calculators can be done in C(53, 4) many ways,
[tex]C(53, 4)= \frac{53!}{4!*49!}= \frac{53*52*51*50}{4*3*2*1}= 292,825[/tex]
4. so, P(none of the selected calculators is defective)=[tex] \frac{52,360}{292,825} =0.18[/tex]
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
