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George Mason University

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a. There are 35 candies total, of which 6 are snickers. We want exactly 3 snickers from
the jar, so the probability is:
However, this expression only represents the probability that the first 3 are snickers
and the last is not. So, we just need to multiply this probability by 4 to get our final
b. Of the 35 candies, 26 are not cadburys. So, the probability that all 4 candies are not
We don't need to take into account the order. So, the final answer is:
c. We will first find the probability that there are no butterfingers, then subtract that
probability from 1 to get the probability that there is at least one butterfinger. Since
22 of the 35 candies are not butterfingers, the probability that there are no
butterfingers is:
The probability that there are no butterfingers is thus . We subtract this quantity
from 1 to get the final answer: .
d. We will first find the probability that there are either exactly 3 or 4 hersheys, then
subtract that probability from 1 to find the probability that there are at most 2
hersheys. We know from part (a) that the probability that there are exactly 3
hersheys is . We can compute the probability that there are exactly 4 hersheys as
follows:
This evaluates to , and thus the probability there are either 3 or 4 hersheys is .
Subtracting this quantity from 1 gives the probability that there are at most 2
hersheys:
e. We already found the probability that there are no butterfingers in part (c). The

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a. There are 35 candies total, of which 6 are snickers. We want exactly 3 snickers from the jar, so the probability is: However, this expression only represents the probability that the first 3 are snickers and the last is not. So, we just need to multiply this probability by 4 to get our final answ ...
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