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, then 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
a.) Wow, 8 defective phones in a batch? That's really pushing the boundaries of "bad luck"! Let's calculate the probability that the entire batch will need testing.
First, we need to consider the probability that at least one defective phone is selected in the initial random sample of 5 phones. To do that, we can use the complementary probability approach. The probability of not selecting a defective phone in one trial is (42/50), so the probability of not selecting a defective phone in all 5 trials is (42/50)^(5). Therefore, the probability of selecting at least one defective phone is 1 - (42/50)^(5).
Since we know that at least one defective phone is selected, the entire batch will need testing. So the probability that the entire batch will need testing is equal to the probability of selecting at least one defective phone, which is 1 - (42/50)^(5).
b.) Two defective phones in a batch? Well, that's still not great, but certainly better than 8!
Using the same reasoning as before, the probability of selecting at least one defective phone in the initial random sample of 5 phones is 1 - (48/50)^(5). And since at least one defective phone is selected, the entire batch will need testing, so the probability that the entire batch will need testing is 1 - (48/50)^(5).
So there you have it! The probability that the entire batch will need testing with 8 defective phones is 1 - (42/50)^(5), and with 2 defective phones is 1 - (48/50)^(5). Good luck in your phone manufacturing adventures!
To find the probability that the entire batch will need testing, we need to determine the probability that at least one defective phone will be selected in the initial random sample.
a) If there are 8 defective phones in the batch:
The probability of selecting at least one defective phone in the initial random sample can be calculated using the complement rule.
The probability of selecting no defective phones in the initial random sample is:
P(no defective phones) = C(50-8, 5)/ C(50, 5) = C(42, 5)/ C(50, 5)
Thus, the probability of selecting at least one defective phone in the initial random sample is:
P(at least one defective phone) = 1 - P(no defective phones)
P(at least one defective phone) = 1 - C(42, 5)/ C(50, 5)
b) If there are 2 defective phones in the batch:
Similarly, for this case, the probability of selecting at least one defective phone in the initial random sample is:
P(at least one defective phone) = 1 - C(48, 5)/ C(50, 5)
Note that the values in the calculations represent the combination formula C(n, r), which calculates the number of ways to choose r items from a set of n items without regard to order.
Please let me know if anything is unclear or if you need further assistance!
To find the probability that the entire batch will need testing, we need to consider two cases:
a) When the batch contains 8 defective phones:
In this case, since at least one of the selected 5 phones is defective, we know that the entire batch will need testing.
Therefore, the probability that the entire batch will need testing is 1.
b) When the batch contains 2 defective phones:
In this case, we need to find the probability that none of the selected 5 phones is defective, which would result in no further testing.
The probability of selecting a non-defective phone from a batch with 2 defective phones is given by:
P(selecting a non-defective phone) = 1 - P(selecting a defective phone)
P(selecting a non-defective phone) = 1 - (Number of defective phones / Total number of phones)
P(selecting a non-defective phone) = 1 - (2 / 50) = 1 - 0.04 = 0.96
Since each of the 5 phones is randomly selected, the probability that none of the selected phones are defective is the probability of selecting a non-defective phone raised to the power of 5 (since we need all 5 selected phones to be non-defective):
P(no selected phones are defective) = (0.96)^5
Finally, to find the probability that the entire batch will need testing, we subtract the probability that none of the selected phones are defective from 1:
P(entire batch needs testing) = 1 - P(no selected phones are defective)
P(entire batch needs testing) = 1 - (0.96)^5
So in the case where the batch contains 2 defective phones, the probability that the entire batch will need testing is approximately 0.184 (or 18.4%).