Due to a manufacturing error, eighteight cans of regular soda were accidentally filled with diet soda and placed into a 2424-pack. suppose that two cans are randomly selected from the 2424-pack. complete parts (a) through (c). (a) determine the probability that both contain diet soda. p(both diet)equals= nothing (round to four decimal places as needed.) (b) determine the probability that both contain regular soda. p(both regular)equals= nothing (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)equals= nothing (round to four decimal places as needed.)

A) the probability would be 0.1014
b) the probability would be 0.4348; no, this would not be unusual
c) the probability would be 0.2319

Explanation
There are 16/24 cans that are regular soda and 8/24 cans that are diet.
a) The probability that the first can is diet would be 8/24.  The probability that the second one is diet after the first one was diet would be 7/23.  The combined probability would be
8/24(7/23) = 56/552 = 0.1014.

b) The probability that the first can is regular would be 16/24.  The probability that the second one is regular would be 15/23.  The combined probability would be
16/24(15/23) = 240/552 = 0.4348.  Since the probability is nearly half, this would not be considered unusual.

c) The probability that the first can is diet would be 8/24.  The probability that the second one is regular after the first can was diet would be 16/23.  This probability would be 
8/24(16/23) = 128/552 = 0.2319.

If we reversed these, and considered getting a regular can first:
The probability of the first can being regular is 16/24.  The probability that the second can is diet after the first one was regular would be 8/23.  The probability would be
16/24(8/23) = 128/552 = 0.2319. The probability for c is the same either way.

The answer for this is 9/1253decimal form 0.007