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 separately: one when the batch contains 8 defective phones and the other when the batch contains 2 defective phones.
Case 1: Batch contains 8 defective phones.
In this case, we have 50 total phones, out of which 8 are defective and 42 are non-defective. We randomly select 5 phones from the batch. There are two possibilities:
a) None of the selected phones is defective:
The probability of selecting a non-defective phone in the first draw is (42/50).
The probability of selecting a non-defective phone in the second draw is (41/49).
The probability of selecting a non-defective phone in the third draw is (40/48).
The probability of selecting a non-defective phone in the fourth draw is (39/47).
The probability of selecting a non-defective phone in the fifth draw is (38/46).
Hence, the probability of none of the selected phones being defective is (42/50) * (41/49) * (40/48) * (39/47) * (38/46).
b) At least one of the selected phones is defective:
The probability of at least one defective phone among the selected phones is 1 - probability of none of the selected phones being defective.
Hence, the probability of at least one defective phone among the selected phones is 1 - [(42/50) * (41/49) * (40/48) * (39/47) * (38/46)].
Case 2: Batch contains 2 defective phones.
In this case, we have 50 total phones, out of which 2 are defective and 48 are non-defective. We randomly select 5 phones from the batch. There are two possibilities:
a) None of the selected phones is defective:
The probability of selecting a non-defective phone in the first draw is (48/50).
The probability of selecting a non-defective phone in the second draw is (47/49).
The probability of selecting a non-defective phone in the third draw is (46/48).
The probability of selecting a non-defective phone in the fourth draw is (45/47).
The probability of selecting a non-defective phone in the fifth draw is (44/46).
Hence, the probability of none of the selected phones being defective is (48/50) * (47/49) * (46/48) * (45/47) * (44/46).
b) At least one of the selected phones is defective:
The probability of at least one defective phone among the selected phones is 1 - probability of none of the selected phones being defective.
Hence, the probability of at least one defective phone among the selected phones is 1 - [(48/50) * (47/49) * (46/48) * (45/47) * (44/46)].
Therefore, the overall probability that the entire batch will need testing is the sum of probabilities from both cases.
For the batch with 8 defective phones: Probability = [1 - (42/50) * (41/49) * (40/48) * (39/47) * (38/46)]
For the batch with 2 defective phones: Probability = [1 - (48/50) * (47/49) * (46/48) * (45/47) * (44/46)]
To calculate the probability, substitute the values into the respective equations.