Question

In a box of assorted cookies, 33% contain chocolate and 12% contain nuts. Of those, 9% contain both chocolate and nuts. A. Find the probability that a cookie contains chocolate or nuts? B. Find the probability that a cookie does not contain chocolate or nuts? Show all work.

Answer 1

Let X=cookies containing chocolate but no nuts, Y=cookies containing nuts but no chocolate, and Z=cookies with both. Let W=cookies containing neither.

These are all percentages.

W+X+Y+Z=100

X+Z=33, Y+Z=12, Z=9 so X=24 and Y=3. And W=100-(24+3+9)=64.

X+Y=27.

Therefore, (A) choc or nuts=27% probability (B) neither choc or nuts=64% probability.