The manager of Toy World has decided to accept a shipment of electronic games if there is no more than 1 defective electronic game in a random sample of size 17. What is the probability that he will accept the shipment if 15% of the electronic games is defective? (Round your answer to three decimal places.)
Let me know your thinking on this one.
To solve this problem, we can use the binomial probability formula. The formula is:
P(x) = C(n, x) * p^x * (1 - p)^(n - x)
Where:
P(x) is the probability of getting exactly x successes
C(n, x) is the combination formula that calculates the number of ways to choose x items from a set of n items
p is the probability of success in a single trial
n is the total number of trials
x is the number of successes
In this case, the manager wants to accept the shipment if there is no more than 1 defective game in a sample of 17, and 15% of the games are defective. So, p = 0.15, n = 17, and x = 0 or x = 1.
Let's calculate the probability for each case:
For x = 0:
P(0) = C(17, 0) * 0.15^0 * (1 - 0.15)^(17 - 0)
C(17, 0) = 1 (since choosing 0 items from a set of n items always results in 1)
0.15^0 = 1 (any number raised to the power of 0 is 1)
(1 - 0.15)^(17 - 0) = 0.85^17 (calculating the probability of not getting a defective game)
P(0) = 1 * 1 * 0.85^17
For x = 1:
P(1) = C(17, 1) * 0.15^1 * (1 - 0.15)^(17 - 1)
C(17, 1) = 17 (since there are 17 ways to choose 1 item from a set of 17 items)
0.15^1 = 0.15 (the probability of getting exactly 1 defective game)
(1 - 0.15)^(17 - 1) = 0.85^16 (calculating the probability of not getting a defective game)
P(1) = 17 * 0.15 * 0.85^16
To find the probability that he will accept the shipment, we need to calculate the sum of P(0) and P(1):
P = P(0) + P(1)
P = (1 * 1 * 0.85^17) + (17 * 0.15 * 0.85^16)
Now, let's calculate this probability.