In a shipment of 20 computer 3 are defective three computers are randomly selected and tested what is the probability that all three are defective if the first and second ones are not replaced after being tested?

What is

(3/20)(2/19)(1/(18) ?

The answer would be 1/1140

To calculate the probability that all three selected computers are defective, we need to consider the fact that the first and second computers are not replaced after being tested.

To start, let's determine the probability of selecting a defective computer from the shipment:

Probability of selecting a defective computer from the shipment
= Number of defective computers / Total number of computers
= 3 / 20

Now, since the first computer is not replaced after being tested, the probability of the second computer being defective will depend on the number of defective computers left in the shipment after the first selection:

Probability of the second computer being defective
= Number of remaining defective computers / Remaining number of computers
= (3 - 1) / (20 - 1)

After the second computer is tested and not replaced, the probability of the third computer being defective will again depend on the remaining number of defective computers in the shipment:

Probability of the third computer being defective
= Number of remaining defective computers / Remaining number of computers
= (3 - 2) / (20 - 2)

Now we need to calculate the probability that all three selected computers are defective, which requires multiplying the individual probabilities together:

Probability that all three selected computers are defective
= Probability of the first computer being defective * Probability of the second computer being defective * Probability of the third computer being defective
= (3/20) * (2/19) * (1/18)

Evaluating this expression, we find:

Probability that all three selected computers are defective
= (3/20) * (2/19) * (1/18)
= 1/1140

Therefore, the probability that all three selected computers are defective is 1 in 1140, or approximately 0.000877, which is equivalent to about 0.0877%.