Question

In a box of assorted cookies, 36% contain chocolate and 12% contain nuts. Of those, 8% contain both chocolate and nuts. Sean is allergic to both chocolate and nuts.

Find the probability that a cookie contains chocolate or nuts (he can’t eat it). (Round to two decimal places)

Answer 1

To find the probability that a cookie contains chocolate or nuts, we will use the principle of inclusion-exclusion for probabilities.

Let:
- \( P(C) \) = Probability that a cookie contains chocolate = 36% = 0.36
- \( P(N) \) = Probability that a cookie contains nuts = 12% = 0.12
- \( P(C \cap N) \) = Probability that a cookie contains both chocolate and nuts = 8% = 0.08

Using the inclusion-exclusion principle, the probability that a cookie contains chocolate or nuts is given by:

\[
P(C \cup N) = P(C) + P(N) - P(C \cap N)
\]

Substituting the values:

\[
P(C \cup N) = 0.36 + 0.12 - 0.08
\]

Calculate \( P(C \cup N) \):

\[
P(C \cup N) = 0.36 + 0.12 - 0.08 = 0.40
\]

Now, rounding to two decimal places, the probability that a cookie contains chocolate or nuts is:

\[
\boxed{0.40}
\]