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)=0.0392 (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 number of regular - soda cans

There are 18 cans in total and 4 are diet - soda cans. So the number of regular - soda cans is $18 - 4=14$.

Step2: Calculate probability of first can being regular

The probability that the first can selected is regular is $\frac{14}{18}$.

Step3: Calculate probability of second can being regular given first is regular

After one regular can is selected, there are 13 regular cans left out of 17 remaining cans. So the probability that the second can is regular given the first is regular is $\frac{13}{17}$.

Step4: Calculate probability that both are regular

By the multiplication rule for independent events (in the context of sequential selection without replacement), $P(\text{both regular})=\frac{14}{18}\times\frac{13}{17}=\frac{182}{306}\approx0.5951$.

Answer:

$0.5951$