Pls help :) grades are due friday i'm scared :(
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There i hope this helped
<span>1.
P(</span>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
while, the total number selections of 4 out of 18+35=53 calculators can be done in C(53, 4) many ways,
4. so, P(none of the selected calculators is defective)=
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
P(</span>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
while, the total number selections of 4 out of 18+35=53 calculators can be done in C(53, 4) many ways,
4. so, P(none of the selected calculators is defective)=
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