5 selected from batch of 50 produced, if one is found defective, than each is tested individually. Find the probability that the entire batch will need testing if the batch contains:
8 defective phones
2 defective phones
To find the probability that the entire batch will need testing, we need to consider two cases: when 8 defective phones are present in the batch and when 2 defective phones are present in the batch.
Case 1: 8 defective phones in the batch
When there are 8 defective phones in the batch, we already know that one phone is defective. Therefore, the remaining 4 phones need to be tested individually.
The probability that the first phone selected from the batch is defective is 8/50.
The probability that the second phone selected from the batch is defective, given that the first phone was defective, is 7/49.
The probability that the third phone selected from the batch is defective, given that the first two phones were defective, is 6/48.
The probability that the fourth phone selected from the batch is defective, given that the first three phones were defective, is 5/47.
The probability that the fifth phone selected from the batch is defective, given that the first four phones were defective, is 4/46.
To find the probability that all 5 phones need testing, we multiply the individual probabilities together:
(8/50) * (7/49) * (6/48) * (5/47) * (4/46) ≈ 0.00040
Therefore, when there are 8 defective phones in the batch, the probability that the entire batch will need testing is approximately 0.00040.
Case 2: 2 defective phones in the batch
When there are 2 defective phones in the batch, and one of them is found to be defective, it means the other 3 phones need to be tested individually.
The probability that the first phone selected from the batch is defective is 2/50.
The probability that the second phone selected from the batch is defective, given that the first phone was defective, is 1/49.
The probability that the third phone selected from the batch is defective, given that the first two phones were defective, is 1/48.
The probability that the fourth phone selected from the batch is defective, given that the first three phones were defective, is 1/47.
The probability that the fifth phone selected from the batch is defective, given that the first four phones were defective, is 1/46.
To find the probability that all 5 phones need testing, we multiply the individual probabilities together:
(2/50) * (1/49) * (1/48) * (1/47) * (1/46) ≈ 0.0000000083
Therefore, when there are 2 defective phones in the batch, the probability that the entire batch will need testing is approximately 0.0000000083.
In summary, the probability that the entire batch will need testing when there are 8 defective phones is approximately 0.00040, and when there are 2 defective phones is approximately 0.0000000083.