Explanation:
probabilities are always
desired cases / totally possible cases
24 cans in total, 4 of them are diet, therefore 20 of them are regular.
2 cans are picked.
for the first can
the probability to pick a diet can is
4/24 = 1/6
the probability to pick a regular is
20/24 = 5/6
a)
to pick 2 diets, the first one has to be diet.
that leaves 23 cans with 3 being diet.
so, the probability for the second can being diet too is
3/23.
as combined event the probability to pull 2 diet cans is
1/6 × 3/23 = 1/2 × 1/23 = 1/46 = 0.02173913...
≈ 0.0217
b)
now the same for picking 2 regular cans.
the first one has to be regular :
5/6
that leaves for the second one 23 cans and 19 of them regular
19/23
combined this gives the probability
5/6 × 19/23 = 95/138 = 0.688405797...
≈ 0.6884
this is not unusual.
what would be considered "usual" or "unusual" ?
c)
one of them is diet, and one of them is regular.
the simple answer ?
it is the only other option, when the cans are not of the same type. it is the opposite of the sum of a) and b).
so, it is
1 - 0.688405797... - 0.02173913 = 0.289855072...
≈ 0.2899
but what to do, if we want to calculate it directly ?
it is the case that either
1. the first can is diet and the second can is regular,
or
2. the first can is regular and the second can is diet.
case 1.
the first can is diet = 1/6
leaving 23 cans with 20 being regular
the second can being regular = 20/23
together
1/6 × 20/23 = 1/3 × 10/23 = 10/69 = 0.144927536...
case 2.
the first can is regular = 5/6
leaving 23 cans with 4 diets.
the second can being diet = 4/23
together
5/6 × 4/23 = 5/3 × 2/23 = 10/69 = 0.144927536...
in an "exclusive or" situation we can add the probabilities.
so, the probability of having exactly one can diet and exactly one can regular is
0.144927536... + 0.144927536... = 0.289855072...
≈ 0.2899
we were correct in the first place !