Can someone please help me

Mathematics

1 answer:

anastassius [24]3 years ago

6 0

Answer:

Step-by-step explanation:

Dont work sorry

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Step-by-step explanation:

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A sample of 4 different calculators is randomly selected from a group containing 18 that are defective and 35 that have no defec

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

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

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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

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3 years ago