Question

Due to a manufacturing error, eight cans of regular soda were accidentally filled with diet soda and placed into a 36-pack. Suppose that two cans are randomly selected from the 36-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.,)

Answer 1

Ans(a) Total diet can in the pack of 36 is 8. And we have to find the probability that if 2 can selected, then both are diet cans.

Out 8 diet soda can 2 can can be drawn in, $\binom{8}{2}=\frac{8!}{6!2!}=\frac{8*7*6!}{6!2!}=28$

And out of 36 pack of cans 2 can drawn in, $\binom{36}{2}=\frac{36!}{34!2!}=\frac{36*35*34!}{2!*34!}=630$

$P(\textrm{Both\:diet})=\frac{28}{630}=0.044444444\approx 0.0444$