Question

You order seventeen burritos to go from a Mexican restaurant, seven with hot peppers and ten without. However, the restaurant forgot to label them. If you pick three burritos at random, find the probability of the given event. (Round your answer to three decimal places.)
At least two have hot peppers.

Answer 1

Answer:

$P=0.369$

Step-by-step explanation:

In this binomial probability case, the success is to pick a burrito with hot peppers, there are 7 out of 17, the probability is p=7/17=0.4118. The formula is

$P=_nC_kp^k(1-p)^{n-k}$

Where n is the number of burritos taken at random, k is the number of success (burrito with hot peppers), p is the probability of success. $_nC_k$ denotes the number of combinations and its formula is:

$_nC_k=\frac{n!}{k!(n-k)!}$

At least two burritos have hot peppers means having two or three burritos, these two cases are calculated and sum to get the probability of at least two burritos that have hot peppers.

$P=_nC_kp^k(1-p)^{n-k}=\frac{n!}{k!(n-k)!}p^k(1-p)^{n-k}$

$P_2=\frac{3!}{2!(3-2)!}(0.4118)^2(1-0.4118)^{3-2}=\frac{3\times2!}{2!1!} (0.1696)(0.5882)^1\\P_2=3 (0.1696)(0.5882)=0.2993$

$P_2=\frac{3!}{3!(3-3)!}(0.4118)^3(1-0.4118)^{3-3}=\frac{1}{0!} (0.0698)(0.5882)^0\\P_2=1 (0.0698)(1)=0.0698$

$P=P_2+P_3=0.2993+0.0698=0.3691$

Round to three decimal places: P=0.369