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Vilka [71]
3 years ago
7

A sample of 4 different calculators is randomly selected from a group containing 19 that are defective and 36 that have no defec

ts. what is the probability that at least one of the 4 calculators in the sample is defective?
Mathematics
1 answer:
Liono4ka [1.6K]3 years ago
6 0
<span><span>The probability of all calculators having no defects is 16.8%. The probability of at least one defect is the complement of that probability. 
ie an 83.2% probability of at least one defect.</span><span>
</span></span>
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only three shapes are mentioned.

But anyhow, 4 shapes allow the following reasoning...

you can start lining up with any of the for shapes, so you have 4 possibilities, for the second position you have 3 possibilities, for the third 2 and for the last only 1.

so 4*3*2*1= 24 possible ways

but with the added rule to keep the circle and the square apart, we have to reduce the number.

by the same step-by-step reasoning we get

4 possibilities for the 1st item

only 2 possibilities for the 2nd item

(if 1st is square or circle, the other one's not an option, if it's not we have to choose one of them to not place them both at point 3 and 4)

only 1 choice for the 3rd item

(reasoning as above, if 2nd is square or circle, we have to the only other, if it isn't, we can put it here at 3rd position)

plus scenarios where either square or circle are on 1st point, the other's on 4th.

I guess that makes 14 possible arrangements

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