Question

due to a manufacturing error, two cans of regular soda were accidentally filled with diet soda and placed alongside cans of regular soda in a 24 - pack. suppose that two cans are randomly selected from the 24 - pack. complete parts (a) through (c).
(a) determine the probability that both contain diet soda.
p(both diet) = 0.0085 (round to four decimal places as needed.)
(b) determine the probability that both contain regular soda.
p(both regular) = 802 (round to four decimal places as needed.)

Explanation:

Step1: Calculate the probability for part (a)

We use the combination formula \(C(n,k)=\frac{n!}{k!(n - k)!}\). Here \(n = 24\) (total number of cans in the pack), and we want to choose \(k = 2\) cans. The number of ways to choose 2 cans out of 24 is \(C(24,2)=\frac{24!}{2!(24 - 2)!}=\frac{24\times23}{2\times1}=276\).
The number of ways to choose 2 diet - soda cans (assuming there are 2 diet - soda cans mistakenly placed). The number of ways to choose 2 diet - soda cans out of 2 is \(C(2,2) = 1\).
The probability \(P(\text{both diet})=\frac{C(2,2)}{C(24,2)}=\frac{1}{276}\approx0.0036\)

Step2: Calculate the probability for part (b)

The number of ways to choose 2 regular - soda cans. The number of regular - soda cans is \(24 - 2=22\). The number of ways to choose 2 regular - soda cans out of 22 is \(C(22,2)=\frac{22!}{2!(22 - 2)!}=\frac{22\times21}{2\times1}=231\)
The probability \(P(\text{both regular})=\frac{C(22,2)}{C(24,2)}=\frac{231}{276}=\frac{77}{92}\approx0.8370\)

Answer:

(a) \(0.0036\)
(b) \(0.8370\)