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Due to a manufacturing error, four cana of regular soda were accidentally filled with diet soda and placed into a 12-pack. Suppose that two cans are randomly selected from the 12-pack. Complete parts (6) through (C)(*) Determine the probability that both contain diet soda,P both diet)(Round to four decimal places as needed)(1) Determine the probability that both contain regular soda,P both regular) - (Round to four decimal places as needed.)Would this be unusual?Ο Νο0 Yes() Determine the probability that exactly one is diet and exactly one is regularP(one diet and one regular) - (Round to four decimal places as needed)

Due to a manufacturing error four cana of regular soda were accidentally filled with diet soda and placed into a 12pack Suppose that two cans are randomly selec class=


Answer :

Total cans = 12

Cans of diet soda, D= 4

Cans of regular soda, R = 8

Two cans are randomly selected from the 12pack:

a) The probability that both contain diet soda:

[tex]\begin{gathered} P(D_1)\times P(D_2|D_1) \\ =\frac{4}{12}\times\frac{3}{11}=0.09090909\approx0.0909(nearest\text{ 4 decimal places)} \end{gathered}[/tex]

Hence, the answer is

[tex]0.0909[/tex]

b)The probability that both contain regular soda:

[tex]\begin{gathered} P(R_1)\times P(R_2|R_1) \\ =\frac{8}{12}\times\frac{7}{11}=0.42424242\approx0.4242(nearest\text{ 4 decimal places)} \end{gathered}[/tex]

Hence, the answer is

[tex]0.4242[/tex]

Therefore, as the probability is more than 0.05. So, it is not unusual.

c) The probability that exactly one is diet and exactly one is regular:

[tex]\begin{gathered} P(D_1)\times P(R_2|D_1)+P(R_1)\times P(D_2|R_1) \\ =\frac{4}{12}\times\frac{8}{11}+\frac{8}{12}\times\frac{4}{11}=0.2424+0.2424=0.4848 \end{gathered}[/tex]

Therefore, the answer is

[tex]0.4848[/tex]

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