Question

due to a manufacturing error, three cans of regular soda were accidentally filled with diet soda and placed alongside cans of regular soda in a 12 - pack. suppose that two cans are randomly selected from the 12 - pack. complete parts (a) through (c).
(a) determine the probability that both contain diet soda.
p(both diet) = 0.0455 (round to four decimal places as needed.)
(b) determine the probability that both contain regular soda.
p(both regular) = (round to four decimal places as needed.)

Explanation:

Step1: Calculate the number of regular soda cans

There are \(12 - 3=9\) regular - soda cans.

Step2: Use the combination formula for probability

The formula for the probability of choosing \(k\) items from \(n\) items is \(P=\frac{C(n,k)}{C(N,K)}\), where \(C(n,k)=\frac{n!}{k!(n - k)!}\). Here, \(N = 12\) (total number of cans), \(K = 2\) (number of cans selected), \(n=9\) (number of regular - soda cans), and \(k = 2\) (number of regular - soda cans to be selected).

First, calculate \(C(12,2)=\frac{12!}{2!(12 - 2)!}=\frac{12\times11}{2\times1}=66\)

Then, calculate \(C(9,2)=\frac{9!}{2!(9 - 2)!}=\frac{9\times8}{2\times1}=36\)

The probability \(P(\text{both regular})=\frac{C(9,2)}{C(12,2)}\)

Step3: Compute the probability value

\(P(\text{both regular})=\frac{36}{66}\approx0.5455\)

Answer:

\(0.5455\)