A sample of 4 different calculators is randomly selected from a group containing 42 that are defective and 20 that have no defects. What is the probability that all four of the calculators selected are defective? No replacement. Round to four decimal places.

 = approx 0.2006

Step-by-step explanation:

The probability that first 1 randomly selected  calculator is defective is

P(1st defect)= 42/(42+20)=42/62=21/31

If the first calculator is defective the residual number of defective calculators is 42-1=41. The residual total number  number of calculators is 62-1=61

So the probability that second calculator is defected

P(2nd defective)=41/61

If both previous calculators are defective the residual number of defective calculators is 42-2=40.  Total residual number of calculators is 62-2=60

So the probability that third calculator is defected

P(3rd defective)=40/60=2/3

Finally the probability that also fourth calculator is defective is 39/59

P(4th defective)=39/59

The resulted probability that all 4 calculators are defective is

P(all 4 are defective)= P(1st defect)* P(2nd defect) * P(3rd defect)* P(4th defect)=21*41*2*39/(31*61*3*59)=67158/334707=0.200647... = approx 0.2006

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step-by-step explanation:

15 batches are the most you can make

step-by-step explanation: