A survey finds customers are charged incorrectly for 2 out of every 10 items. Suppose a customer purchases 13 items. Find the probability that the customer is charged incorrectly on at most 2 items.The probability that the customer is charged incorrectly on at most 2 items is...
this is a binomial probability ... (c + i)^13 ... sum the 1st three terms
(13C0 * .8^13) + (13C1 * .8^12 * .2 + (13C2 * .8^11 * .2^2)
(0.2)^0 (0.8)^13+
(0.2)^1 (0.8)^12+
(0.2)^2 (0.8)^11
= 0.5017
what about the term coefficients?
To find the probability that the customer is charged incorrectly on at most 2 items, we need to calculate the probability of being charged incorrectly on 0, 1, or 2 items, and then sum up these probabilities.
Let's start by finding the probability of being charged incorrectly on exactly 0 items. Since the survey finds customers are charged incorrectly for 2 out of every 10 items, the probability of being charged incorrectly on any single item is 2/10 or 0.2. Therefore, the probability of being charged incorrectly on exactly 0 items is the complement of that, which is 1 - 0.2, or 0.8.
Next, let's find the probability of being charged incorrectly on exactly 1 item. Again, the probability of being charged incorrectly on a single item is 0.2. Since the customer purchased 13 items, there are multiple combinations of which item can be charged incorrectly. We can calculate this probability using the binomial probability formula:
P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)
For k = 1:
P(X = 1) = C(13, 1) * (0.2)^1 * (1 - 0.2)^(13 - 1)
= 13 * 0.2 * 0.8^12
≈ 0.2824
Finally, let's find the probability of being charged incorrectly on exactly 2 items. Again, using the binomial probability formula:
P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)
For k = 2:
P(X = 2) = C(13, 2) * (0.2)^2 * (1 - 0.2)^(13 - 2)
= (13! / (2! * (13-2)!)) * 0.2^2 * 0.8^11
≈ 0.3025
Now, to find the probability of being charged incorrectly on at most 2 items, we need to sum up the probabilities of being charged incorrectly on exactly 0, 1, or 2 items:
P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)
= 0.8 + 0.2824 + 0.3025
≈ 0.3849
Therefore, the probability that the customer is charged incorrectly on at most 2 items is approximately 0.3849 or 38.49%.