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) = (round to four decimal places as needed.)

Explanation:

Step1: Calculate the probability of selecting the first diet - soda can

The total number of cans is \(n = 12\), and the number of diet - soda cans is \(m=3\). The probability of selecting the first diet - soda can is \(P_1=\frac{3}{12}\).

Step2: Calculate the probability of selecting the second diet - soda can

After selecting one diet - soda can, the number of remaining cans is \(n_1 = 11\), and the number of remaining diet - soda cans is \(m_1=2\). The probability of selecting the second diet - soda can given that the first one was diet is \(P_2=\frac{2}{11}\).

Step3: Calculate the probability that both are diet - soda cans

By the multiplication rule for dependent events \(P(\text{both diet})=P_1\times P_2\). Substitute \(P_1=\frac{3}{12}\) and \(P_2 = \frac{2}{11}\) into the formula:

$$
LATEXBLOCK0
$$

Answer:

\(0.0455\)