A cellular phone manufacturer randomly selects 5 phones from every batch of 50 produced. If at least one of the phones is found to be defective than each phone in the batch is tested individually. Find the probability that the entire batch will need testing if the batch contains
a). 8 defective phones
b). 2 defective phones
To find the probability that the entire batch will need testing, we need to consider the probability that at least one defective phone is selected from the batch. Let's calculate the probability for each case:
a) Batch contains 8 defective phones:
In this case, there are 50 phones in the batch and 8 of them are defective. To find the probability that at least one defective phone is selected, we can calculate the probability that all 5 randomly selected phones are non-defective and subtract it from 1.
The probability that a phone is non-defective is given by:
P(non-defective) = (total non-defective phones) / (total phones)
= (50 - 8) / 50
= 0.84
The probability that all 5 phones selected are non-defective is:
P(all non-defective) = P(non-defective) * P(non-defective) * P(non-defective) * P(non-defective) * P(non-defective)
= 0.84 * 0.84 * 0.84 * 0.84 * 0.84
≈ 0.413
Therefore, the probability that at least one defective phone is selected and the entire batch needs testing is:
P(testing needed) = 1 - P(all non-defective)
= 1 - 0.413
≈ 0.587 (or 58.7%)
b) Batch contains 2 defective phones:
In this case, there are 50 phones in the batch and 2 of them are defective. Similar to the previous case, we can calculate the probability that all 5 randomly selected phones are non-defective and subtract it from 1.
The probability that a phone is non-defective is given by:
P(non-defective) = (total non-defective phones) / (total phones)
= (50 - 2) / 50
= 0.96
The probability that all 5 phones selected are non-defective is:
P(all non-defective) = P(non-defective) * P(non-defective) * P(non-defective) * P(non-defective) * P(non-defective)
= 0.96 * 0.96 * 0.96 * 0.96 * 0.96
≈ 0.814
Therefore, the probability that at least one defective phone is selected and the entire batch needs testing is:
P(testing needed) = 1 - P(all non-defective)
= 1 - 0.814
≈ 0.186 (or 18.6%)