In the production of a particular model of cell phones, it is found that 5% are defective. Cellular phones are selected at random and put into boxes of 10. If a box is selected at random, find the probability that it contains.
(i) Exactly three defective phones
(ii) Less than four defective phones
(iii)Now if two boxes are randomly selected, find the probability that there are, no defective phones in either box.
To solve these problems, we can use the binomial probability formula:
P(X = k) = C(n, k) * p^k * (1-p)^(n-k)
Where:
P(X = k) is the probability of getting exactly k successes,
n is the total number of trials,
k is the number of successful outcomes,
p is the probability of success in a single trial, and
C(n, k) is the binomial coefficient, which is computed as n! / (k! * (n - k)!)
(i) To find the probability of exactly three defective phones in a randomly selected box of 10, we can use the binomial probability formula:
P(X = 3) = C(10, 3) * 0.05^3 * (1-0.05)^(10-3)
Using a calculator, we can find:
P(X = 3) ≈ 0.2389
So the probability that a randomly selected box contains exactly three defective phones is approximately 0.2389.
(ii) To find the probability of less than four defective phones, we need to find the sum of the probabilities of getting 0, 1, 2, or 3 defective phones.
P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
Using the binomial probability formula for each value, and then adding them up, we get:
P(X < 4) ≈ 0.9876
So the probability that a randomly selected box contains less than four defective phones is approximately 0.9876.
(iii) To find the probability that there are no defective phones in either of the two randomly selected boxes, we need to find the product of the probabilities of no defective phones in the first box and no defective phones in the second box.
Using the binomial probability formula for each box, we get:
P(no defective phones in Box 1) = P(X = 0) = C(10, 0) * 0.05^0 * (1-0.05)^(10-0)
P(no defective phones in Box 2) = P(X = 0) = C(10, 0) * 0.05^0 * (1-0.05)^(10-0)
Calculating these probabilities, we find:
P(no defective phones in Box 1) ≈ 0.5987
P(no defective phones in Box 2) ≈ 0.5987
To find the probability of both events happening (no defective phones in either box), we multiply the probabilities:
P(no defective phones in either box) = P(no defective phones in Box 1) * P(no defective phones in Box 2)
P(no defective phones in either box) ≈ 0.3582
So the probability that there are no defective phones in either of the two randomly selected boxes is approximately 0.3582.
To find the probability in each scenario, we can use the concept of binomial probability.
(i) To find the probability of exactly three defective phones in a box of 10, we can use the binomial probability formula: P(X = k) = (n C k) * p^k * (1-p)^(n-k), where n is the number of trials, k is the number of successful outcomes, p is the probability of success, and (n C k) denotes the number of combinations.
In this case, n = 10 (as we are selecting from a box of 10 phones), k = 3, and p = 5% = 0.05 (as the probability of selecting a defective phone is 5%). Therefore, the probability can be calculated as follows:
P(X = 3) = (10 C 3) * 0.05^3 * (1-0.05)^(10-3)
To calculate (10 C 3), which represents the number of combinations of selecting 3 defective phones from 10, we can use the formula: (n C k) = n! / (k! * (n-k)!)
(10 C 3) = 10! / (3! * (10-3)!) = 10! / (3! * 7!)
Using this information, the probability of exactly three defective phones can be calculated.
(ii) To find the probability of less than four defective phones, we need to calculate the probabilities of having 0, 1, 2, and 3 defective phones, and then sum them up:
P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)
Using the binomial probability formula as explained above, we can calculate each individual probability and sum them up to get the final result.
(iii) To find the probability of no defective phones in either box, we need to consider the probability of no defective phones in the first box and no defective phones in the second box.
Let's denote the probability of no defective phones in a box as P(N) and the probability of defective phones as P(D). Since 5% of the phones are defective, P(N) = 1 - P(D) = 1 - 0.05 = 0.95.
To find the probability of no defective phones in either box, we multiply the two probabilities together:
P(No defective phones in either box) = P(N in box 1) * P(N in box 2) = P(N) * P(N) = 0.95 * 0.95
Hence, by multiplying the individual probabilities, we can find the probability of no defective phones in either box.