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 scenarios: one where there are 8 defective phones in the batch, and the other where there are 2 defective phones in the batch.
Scenario 1: Batch contains 8 defective phones
In this scenario, we select 5 phones from the batch. If one of these 5 phones is defective, we will need to test the entire batch to identify all the defective phones.
To calculate the probability, we first need to find the probability of selecting at least one defective phone out of the 5 phones. Since there are 8 defective phones in the batch and 50 total phones, the probability of selecting at least one defective phone in the first draw is:
P(defective phone in first draw) = (number of defective phones / total number of phones) = 8/50 = 0.16
Since we are interested in at least one defective phone, we subtract this probability from 1 to get the probability of not selecting any defective phones in the first draw:
P(no defective phone in first draw) = 1 - P(defective phone in first draw) = 1 - 0.16 = 0.84
Now, in order for the entire batch to need testing, we need to continue selecting phones until we find a defective phone. The probability of selecting all non-defective phones in the second, third, fourth, and fifth draws can be calculated as:
P(no defective phone in second draw) = P(no defective phone in first draw) = 0.84
P(no defective phone in third draw) = P(no defective phone in first draw) = 0.84
P(no defective phone in fourth draw) = P(no defective phone in first draw) = 0.84
P(no defective phone in fifth draw) = P(no defective phone in first draw) = 0.84
Since each draw is independent, we can multiply these probabilities together to get the overall probability:
P(entire batch needs testing | there are 8 defective phones in the batch) = P(no defective phone in second draw) * P(no defective phone in third draw) * P(no defective phone in fourth draw) * P(no defective phone in fifth draw)
= 0.84 * 0.84 * 0.84 * 0.84 = 0.5997
So, the probability that the entire batch will need testing if there are 8 defective phones is approximately 0.5997.
Scenario 2: Batch contains 2 defective phones
In this scenario, the calculations are similar but the starting probabilities change because there are only 2 defective phones in the batch.
P(defective phone in first draw) = (number of defective phones / total number of phones) = 2/50 = 0.04
P(no defective phone in first draw) = 1 - P(defective phone in first draw) = 1 - 0.04 = 0.96
The subsequent calculations for the second, third, fourth, and fifth draws are the same as in Scenario 1. So:
P(entire batch needs testing | there are 2 defective phones in the batch) = 0.96 * 0.96 * 0.96 * 0.96 = 0.8493
Therefore, the probability that the entire batch will need testing if there are 2 defective phones is approximately 0.8493.