Question

(a) determine the probability that both contain diet soda. p(both diet) = 0.0036 (round to four decimal places as needed.) (b) determine the probability that both contain regular soda. p(both regular) = 0.8370 (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 ways to choose 2 cans from 24

The formula for combinations is \(C(n,r)=\frac{n!}{r!(n - r)!}\), where \(n = 24\) and \(r=2\).

$$C(24,2)=\frac{24!}{2!(24 - 2)!}=\frac{24\times23\times22!}{2\times1\times22!}=276$$

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

Let's assume there are \(x\) diet cans and \(y\) regular cans. Since we are choosing 1 diet and 1 regular, the number of ways is \(C(x,1)\times C(y,1)\). But we can also use the formula \(n_1\times n_2\) (where \(n_1\) is the number of diet cans and \(n_2\) is the number of regular cans). Let's assume there are 2 diet and 22 regular (since total is 24). The number of ways to choose 1 diet and 1 regular is \(2\times22 = 44\)

Step3: Calculate the probability \(P(\text{one diet and one regular})\)

The probability formula is \(P=\frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}}\)

$$P=\frac{44}{276}\approx0.1594$$

Answer:

\(0.1594\)