Question

due to a manufacturing error, four cans of regular soda were accidentally filled with diet soda and placed alongside cans of regular soda in an 18 - pack. suppose that two cans are randomly selected from the 18 - 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 probability of first - can being diet

The total number of cans is $n = 18$, and the number of diet - soda cans is $m = 4$. The probability that the first can selected is a diet - soda can is $P_1=\frac{4}{18}$.

Step2: Calculate probability of second - can being diet given first is diet

After one diet - soda can is selected, the number of remaining cans is $n_1 = 17$, and the number of remaining diet - soda cans is $m_1 = 3$. The probability that the second can selected is a diet - soda can given that the first one is a diet - soda can is $P_2=\frac{3}{17}$.

Step3: Calculate probability of both being diet

By the multiplication rule for independent events (in the context of sequential selection without replacement), the probability that both cans are diet - soda cans is $P = P_1\times P_2=\frac{4}{18}\times\frac{3}{17}=\frac{12}{306}\approx0.0392$.

Answer:

$0.0392$