You order twelve burritos to go from a Mexican restaurant, five with hot peppers and seven without. However, the restaurant forgot to label them. If you pick three burritos at random, find the probability that at least two have hot peppers. (Round the answer to three decimal places.)
(I put 0.36 but it was marked wrong).
You order fifteen burritos to go from a Mexican restaurant, eight with hot peppers and seven without. However, the restaurant forgot to label them. If you pick five burritos at random, find the probability of the given event. (Round your answer to three decimal places.)
All have hot peppers.
(5C2 x 7C1) + (5C3) / 12C3 = .3636 = .364
To find the probability that at least two burritos have hot peppers, we can approach this problem using the concept of combinations.
First, let's determine the total number of ways to choose 3 burritos out of 12. This can be calculated using the combination formula:
nCr = n! / (r! * (n-r)!)
In this case, n is the total number of burritos (12) and r is the number of burritos we are selecting (3).
12C3 = 12! / (3! * (12-3)!)
= 12! / (3! * 9!)
= (12 * 11 * 10) / (3 * 2 * 1)
= 220
There are a total of 220 possible combinations of 3 burritos that can be chosen.
Now, let's calculate the number of combinations where at least two burritos have hot peppers. Since there are 5 burritos with hot peppers, we have two cases to consider:
1. Selecting 2 burritos with hot peppers and 1 without.
2. Selecting all 3 burritos with hot peppers.
Case 1: Selecting 2 burritos with hot peppers and 1 without.
To calculate the number of combinations for this case, we need to choose 2 burritos out of 5 with hot peppers and 1 burrito out of 7 without hot peppers.
5C2 * 7C1 = (5! / (2! * (5-2)!) * 7! / (1! * (7-1)!)
= (5 * 4 / (2 * 1)) * (7 / 1)
= 10 * 7
= 70
Case 2: Selecting all 3 burritos with hot peppers.
To calculate the number of combinations for this case, we simply need to choose 3 burritos out of the 5 with hot peppers.
5C3 = 5! / (3! * (5-3)!)
= (5 * 4 * 3) / (3 * 2 * 1)
= 10
Now, let's calculate the total number of combinations that satisfy the condition of at least two burritos having hot peppers by summing the combinations from both cases:
70 + 10 = 80
Therefore, there are 80 combinations where at least two burritos have hot peppers out of the total 220 combinations.
Now, to find the probability, we divide the number of combinations where at least two burritos have hot peppers by the total number of combinations:
Probability = (Number of favorable outcomes) / (Total number of outcomes)
= 80 / 220
= 0.3636 (rounded to four decimal places)
Thus, the probability that at least two burritos have hot peppers is approximately 0.364 or 36.4%.
P -- pepper
N -- without pepper
P(P) = 5/12
P(N) = 7/12
to have at least 2P, you could have 2P or 3P,
Two peppers could be PPN PNP NPP
Prob of that is 3x(5/12)^2(7/12) = .3038
Three peppers = (5/12)^3 = .0723
adding these we get .376