SOLUTION: A box contains 20 DVDs, 4 of which are defective If two DVDs are selected at random(without replacement) from this box, what is the probability that both are defective

Question 1138114: A box contains 20 DVDs, 4 of which are defective If two DVDs are selected at random(without replacement) from this box, what is the probability that both are defective
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!

Define two events
A = selecting a defective DVD on the first selection
B = selecting a defective DVD on the second selection

The probability of event A happening is
P(A) = 4/20 = 1/5
since there are 4 defective DVDs out of 20 total

P(B|A) = probability of event B happening given that event A has occurred already
P(B|A) = probability of getting second defective DVD if the first is known to be defective
P(B|A) = 3/19
No replacement is made. There are 4-1 = 3 defective DVDs left out of 20-1 = 19 overall

Multiply the two probabilities
P(A and B) = P(A)*P(B|A) ...... this is a rearrangement of the conditional probability formula
P(A and B) = (1/5)*(3/19) ..... substitution
P(A and B) = (1*3)/(5*19)
P(A and B) = 3/95
P(A and B) = 0.0316 which is approximate
P(A and B) = 3.16% also approximate

Answer as a fraction: 3/95
Answer in decimal form: 0.0316 (approximate)
Answer in percent form: 3.16% (approximate)

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