You order twelve burritos to go from a Mexican restaurant, five with hot peppers and seven without. However, the restaurant forgot to label them. You pick three burritos at random. Find the probability of the events (as a reduced fraction and as a percent rounded to the nearest tenth of a percent).
a) All 3 of the burritos have hot peppers.
b) None of the three burritos have hot peppers.
c) Exactly one of the three burritos has hot peppers.
d) Exactly two of the three burritos have hot peppers.
e) At least one of the three burritos has hot peppers.
f) At least two of the three burritos have hot peppers.
g) At most two of the three burritos have hot peppers.
If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
P(3) = (5/12)^3
P(1of3) = (5/12)(7/12)^2
e) At least one = one, two or three
Either-or probabilities are found by adding the individual probabilities.
To find the probability of the events, we first need to determine the total number of possible outcomes.
a) All 3 of the burritos have hot peppers:
There are 5 burritos with hot peppers to choose from. Since we are picking three burritos, the probability is (5/12) * (4/11) * (3/10) = 60/660 = 1/11. As a percent, this is approximately 9.1%.
b) None of the three burritos have hot peppers:
There are 7 burritos without hot peppers to choose from. Similar to the previous case, the probability is (7/12) * (6/11) * (5/10) = 210/1,320 = 1/6.29. As a percent, this is approximately 15.9%.
c) Exactly one of the three burritos has hot peppers:
For this case, we have three possible scenarios where exactly one burrito can have hot peppers. We need to consider the different positions of the burrito with hot peppers. The probability is calculated as follows:
(5/12) * (7/11) * (6/10) + (7/12) * (5/11) * (6/10) + (7/12) * (6/11) * (5/10) = 210/1,320 + 210/1,320 + 210/1,320 = 3 * 210/1,320 = 630/1,320 = 9/22. As a percent, this is approximately 40.9%.
d) Exactly two of the three burritos have hot peppers:
Similarly, we have three possible scenarios where exactly two burritos can have hot peppers. The probability is calculated as follows:
(5/12) * (4/11) * (7/10) + (5/12) * (7/11) * (4/10) + (7/12) * (5/11) * (4/10) = 140/1,320 + 140/1,320 + 140/1,320 = 3 * 140/1,320 = 420/1,320 = 7/22. As a percent, this is approximately 31.8%.
e) At least one of the three burritos has hot peppers:
To find this probability, we can consider the complement probability: the probability that none of the three burritos has hot peppers. From part b), we found that the probability of none of the burritos having hot peppers is 1/6.29. Hence, the probability of at least one burrito having hot peppers is 1 - 1/6.29 = 5.29/6.29 ≈ 0.841. As a percent, this is approximately 84.1%.
f) At least two of the three burritos have hot peppers:
We can approach this similarly to part e). The complement probability is the probability that fewer than two burritos have hot peppers. Specifically, it means that either all three have no hot peppers or exactly one has hot peppers. Hence, the probability is (1/6.29) + (9/22) = 311/1,320 ≈ 0.235. As a percent, this is approximately 23.5%.
g) At most two of the three burritos have hot peppers:
To find this probability, we can consider the complement probability: the probability that all three burritos have hot peppers. From part a), we found that the probability of all three burritos having hot peppers is 1/11. Hence, the probability of at most two burritos having hot peppers is 1 - 1/11 = 10/11. As a percent, this is approximately 90.9%.
So, the probabilities as reduced fractions and as percents are:
a) 1/11, 9.1%
b) 1/6.29, 15.9%
c) 9/22, 40.9%
d) 7/22, 31.8%
e) 5.29/6.29, 84.1%
f) 311/1,320, 23.5%
g) 10/11, 90.9%