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) = 0.5948 (round to four decimal places as needed.) would this be unusual? yes no (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 total number of ways to select 2 cans

The number of ways to choose 2 cans out of 18 is given by the combination formula $C(n,r)=\frac{n!}{r!(n - r)!}$, where $n = 18$ and $r=2$. So, $C(18,2)=\frac{18!}{2!(18 - 2)!}=\frac{18\times17}{2\times1}= 153$.

Step2: Calculate number of ways to select 1 diet and 1 regular can

There are 4 diet - soda cans and $18 - 4=14$ regular - soda cans. The number of ways to choose 1 diet can out of 4 is $C(4,1)=\frac{4!}{1!(4 - 1)!}=4$, and the number of ways to choose 1 regular can out of 14 is $C(14,1)=\frac{14!}{1!(14 - 1)!}=14$. By the multiplication principle, the number of ways to choose 1 diet and 1 regular can is $C(4,1)\times C(14,1)=4\times14 = 56$.

Step3: Calculate the probability

The probability $P$ that exactly one is diet and exactly one is regular is the number of favorable outcomes divided by the number of total outcomes. So, $P=\frac{56}{153}\approx0.3660$.

Answer:

$0.3660$