In a shipment of 50 calculators,4 are defective. Two calculators are randomly selected and tested. What is the probability that both are defective if the first one is not replaced after being tested?
4/50 * 3/49
To find the probability that both calculators are defective, given that the first one is not replaced, we can use conditional probability.
Let's break down the problem step-by-step:
Step 1: Calculate the probability of selecting a defective calculator from the shipment.
There are 4 defective calculators out of a total of 50. Therefore, the probability of selecting a defective calculator is 4/50 or 2/25.
Step 2: Calculate the probability of selecting a defective calculator given that the first one is not replaced.
After selecting the first calculator, there will be one less calculator in the shipment, so there will be 49 calculators remaining, with 3 defective calculators. Therefore, the probability of selecting a defective calculator as the second one, given that the first one was not replaced, is 3/49.
Step 3: Multiply the probabilities together to calculate the probability that both calculators are defective.
The probability of both calculators being defective is the product of the probabilities calculated in steps 1 and 2.
(2/25) * (3/49) = 6/1225 or approximately 0.0049
So, the probability that both calculators are defective, given that the first one is not replaced, is approximately 0.0049.
To find the probability that both calculators are defective, given that the first one is not replaced after being tested, we need to use conditional probability.
The probability of the first calculator being defective is 4 out of 50, or 4/50.
After the first calculator is tested and not replaced, there are now 49 calculators left in the shipment, with 3 defective ones.
The probability of the second calculator being defective, given that the first calculator is already known to be defective and not replaced, is 3 out of 49, or 3/49.
To find the probability of both calculators being defective, we multiply the probabilities of each event happening:
P(both defective) = P(first defective) * P(second defective given first defective)
P(both defective) = (4/50) * (3/49)
P(both defective) = 12/2450
Therefore, the probability that both calculators are defective, given that the first one is not replaced after being tested, is 12/2450.