A survey finds customers are charged incorrectly for an item 20% of the time. Suppose a customer purchases 15 items. Find the probability that the customer is charged incorrectly on at most 2 items.
Pr(at most 2)= Pr(none) + Pr(1) + Pr (2)
this is a binomial probability ... (c + i)^15 ... sum the 1st three terms
(15C0 * .8^15) + (15C1 * .8^12 * .2 ) + (15C2 * .8^11 * .2^2)
using the Combination coefficient as nCx.
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 separately, and then sum them up.
First, let's find the probability of being charged correctly for an item, which is 1 - 20% = 80% = 0.80.
Now let's calculate the probability of being charged incorrectly on 0 items:
P(0 incorrect charges) = (0.80)^15 ≈ 0.035
Next, let's calculate the probability of being charged incorrectly on 1 item:
P(1 incorrect charge) = 15C1 * (0.20)^1 * (0.80)^14 ≈ 0.032
Finally, let's calculate the probability of being charged incorrectly on 2 items:
P(2 incorrect charges) = 15C2 * (0.20)^2 * (0.80)^13 ≈ 0.021
Now, we can sum up these probabilities:
P(at most 2 incorrect charges) = P(0 incorrect charges) + P(1 incorrect charge) + P(2 incorrect charges)
= 0.035 + 0.032 + 0.021
≈ 0.088
Therefore, the probability that the customer is charged incorrectly on at most 2 items is approximately 0.088 or 8.8%.
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 items, 1 item, or 2 items and then sum up these probabilities.
Let's calculate each probability separately:
1. Probability of being charged incorrectly on 0 items:
Since the survey found that customers are charged incorrectly 20% of the time, the probability of being charged correctly on one item is 1 - 20% = 80% or 0.8. So, the probability of being charged correctly on all 15 items is 0.8^15. Therefore, the probability of being charged incorrectly on 0 items is (0.8^15) or approximately 0.0351.
2. Probability of being charged incorrectly on 1 item:
To find the probability of being charged incorrectly on 1 item, we need to multiply the probability of being charged incorrectly (20% or 0.2) by the probability of being charged correctly on the remaining 14 items (0.8^14). Therefore, the probability of being charged incorrectly on 1 item is (0.2 * 0.8^14) or approximately 0.0611.
3. Probability of being charged incorrectly on 2 items:
To find the probability of being charged incorrectly on 2 items, we need to calculate the probability of being charged incorrectly on 2 items and correctly on the remaining 13 items. The probability of being charged incorrectly on 2 items is (0.2^2) and the probability of being charged correctly on the remaining 13 items is (0.8^13). Therefore, the probability of being charged incorrectly on 2 items is (0.2^2 * 0.8^13) or approximately 0.0740.
Now, we can sum up these probabilities:
Probability = Probability of being charged incorrectly on 0 items + Probability of being charged incorrectly on 1 item + Probability of being charged incorrectly on 2 items
= 0.0351 + 0.0611 + 0.0740
= 0.1703
Therefore, the probability that the customer is charged incorrectly on at most 2 items is approximately 0.1703 or 17.03%.