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)=(0.5455) (round to four decimal places as needed.)
would this be unusual?
no
yes
(c) determine the probability that exactly one is diet and exactly one is regular.
p(one diet and one regular)=( ) (round to four decimal places as needed.)

Explanation:

Step1: Calculate the number of regular sodas

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

Step2: Use the formula for probability of one - diet and one - regular

The formula for the probability of choosing one diet and one regular is \(P=\frac{C(3,1)\times C(9,1)}{C(12,2)}\)

We know that the combination formula \(C(n,k)=\frac{n!}{k!(n - k)!}\)

First, \(C(3,1)=\frac{3!}{1!(3 - 1)!}=\frac{3!}{1!2!}=3\)

Second, \(C(9,1)=\frac{9!}{1!(9 - 1)!}=\frac{9!}{1!8!}=9\)

Third, \(C(12,2)=\frac{12!}{2!(12 - 2)!}=\frac{12\times11\times10!}{2\times1\times10!}=66\)

Then \(P=\frac{3\times9}{66}=\frac{27}{66}\approx0.4091\)

Answer:

\(0.4091\)