A batch of 445 containers for frozen orange juice contains 3 that are defective.
ID: 3207192 • Letter: A
Question
A batch of 445 containers for frozen orange juice contains 3 that are defective. Two are selected, at random, without replacement from the batch. What is the probability that the second one selected is defective given that the first one was defective? Round your answer to five decimal places (e.g. 98.76543). What is the probability that both are defective? Round your answer to seven decimal places (e.g. 98.7654321). What is the probability that both are acceptable? (e.g. 98.765). Three containers are selected, at random, without replacement, from the batch. What is the probability that the third one selected is defective given that the first and second one selected were defective? (e.g. 98.765). What is the probability that the third one selected is defective given that the first one selected was defective and the second one selected was okay? Round your answer to five decimal places (e.g. 98.76543). What is the probability that all three are defective? (e.g. 98.765).Explanation / Answer
Given,
number of containers=445
no of defectives=3
no of acceptable=445-3=442
A.a) probability of second is defective given first is defective
after first is defective we are left with 2 defective in 444 containers
hence, probability=2C1/444C1
probability=2/444
probability=0.00450
A.b) probability that both are defective
probability=(no of ways of selecting defective)/(no of ways of selecting containers)
probability=3C2/445C2
probability=3/(445*444/2)
probability=0.0000303
A.c) probability that both are acceptable
probability=(no of ways of selecting acceptable)/(no of ways of selecting containers)
probability =442C2/445C2
probability =442*441/445*444
probability =0.986
A.d) probability that third is defective given first and second is defective
after 2 defective we are left with 1 defective containers in the lot
hence probability=3-2/445-2
probability=1/443
probability =0.002
A.e) probability that third is defective given first is defective and second is okay
after selecting first and second containers we are left with
2 defective containers and 441 acceptable containers
hence probability=2C1/443C1
probability=0.00451
A.f) probability that all 3 are defective
probability=3C3/445C3
probability =3*2*1/445*444*443
probability =6/87527940
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