A manufacturer ships toasters in cartons of 25. In each carton, they estimate a 8% chance that one of the toasters will need to be sent back for minor repairs. In a batch of 12,000 toasters, what are the chances that fewer than 950 need to be returned?
To solve this problem, we can use the binomial probability formula, which is given by:
P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)
Where:
P(X = k) is the probability that exactly k successes occur
C(n, k) is the number of combinations of n items taken k at a time
p is the probability of success on a single trial
n is the number of trials
In this case, the number of trials (n) is 12,000, the probability of success (p) is 0.08 (8%), and we want to find the probability that fewer than 950 toasters need to be returned (X < 950).
Let's calculate it step-by-step:
Step 1: Calculate the probability of one toaster needing to be sent back for repairs:
P(returned) = 0.08
Step 2: Calculate the probability that fewer than 950 toasters need to be returned:
P(X < 950) = P(X = 0) + P(X = 1) + P(X = 2) + ... + P(X = 949)
We can use the binomial probability formula to calculate each term:
P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)
P(X = 0) = C(12000, 0) * 0.08^0 * (1 - 0.08)^(12000 - 0)
P(X = 1) = C(12000, 1) * 0.08^1 * (1 - 0.08)^(12000 - 1)
P(X = 2) = C(12000, 2) * 0.08^2 * (1 - 0.08)^(12000 - 2)
...
P(X = 949) = C(12000, 949) * 0.08^949 * (1 - 0.08)^(12000 - 949)
Step 3: Calculate each term and sum them up to find the probability:
P(X < 950) = P(X = 0) + P(X = 1) + P(X = 2) + ... + P(X = 949)
Let's calculate it using a computer program or a calculator.
To solve this problem, we will use the binomial probability formula. The formula is as follows:
P(X = k) = (n C k) * p^k * (1 - p)^(n - k)
Where:
- P(X = k) is the probability of getting exactly k successes,
- n is the number of trials (number of toasters in this case),
- p is the probability of success (probability that a toaster needs repairs),
- (n C k) is the binomial coefficient, which represents the number of ways to choose k successes from n trials,
- k is the number of successes.
In this case, we want to find the probability that fewer than 950 toasters need to be returned, so we need to calculate the cumulative probability of getting 0 to 949 successful (defective) toasters.
First, let's calculate the probability of one toaster needing repairs:
p = 0.08
Next, calculate the number of toasters in the batch:
n = 12,000
Now, we will calculate the cumulative probability of getting 0 to 949 successful (defective) toasters:
P(X < 950) = P(X = 0) + P(X = 1) + ... + P(X = 949)
Using the binomial formula for each value of k, we can calculate the probabilities for each case:
P(X = 0) = (n C 0) * p^0 * (1 - p)^(n - 0)
P(X = 1) = (n C 1) * p^1 * (1 - p)^(n - 1)
...
P(X = 949) = (n C 949) * p^949 * (1 - p)^(n - 949)
Now, we can calculate these probabilities and sum them up to find the final answer.