Suppose there are two full bowls of cookies. Bowl #1 has 10 chocolate chip and 40 oatmeal, while bowl #2 has 30 chocolate chip cookies and 20 oatmeal cookies. Our friend Bob picks a bowl at random, and then picks a cookie at random. We may assume there is no reason to believe Bob treats one bowl differently from another, likewise for the cookies. The cookie turns out to be a oatmeal one. How probable is it that Bob picked it out of Bowl #2?
P(bolw #1) = 1/2
if #1, P(choc) = 10/50
See what you can do with that, knowing what you do about conditional probabilities.
To determine the probability that Bob picked the oatmeal cookie from Bowl #2, we can use Bayes' theorem.
Let's define the following probabilities:
P(B1) - Probability of picking Bowl #1
P(B2) - Probability of picking Bowl #2
P(O|B1) - Probability of picking an oatmeal cookie given that Bowl #1 was picked
P(O|B2) - Probability of picking an oatmeal cookie given that Bowl #2 was picked
We are given the specific values:
P(B1) = 0.5 (since Bob picks a bowl at random, each bowl has an equal chance of being picked)
P(B2) = 0.5
P(O|B1) = 40/(10+40) = 40/50 = 0.8 (since there are 40 oatmeal cookies out of 50 in Bowl #1)
P(O|B2) = 20/(30+20) = 20/50 = 0.4 (since there are 20 oatmeal cookies out of 50 in Bowl #2)
Now, let's calculate the probability that Bob picked the oatmeal cookie from Bowl #2 using Bayes' theorem:
P(B2|O) = (P(O|B2) * P(B2)) / { (P(O|B1) * P(B1)) + (P(O|B2) * P(B2)) }
= (0.4 * 0.5) / { (0.8 * 0.5) + (0.4 * 0.5) }
= 0.2 / (0.4 + 0.2)
= 0.2 / 0.6
= 1/3
≈ 0.3333 or 33.33%
Therefore, the probability that the oatmeal cookie was picked from Bowl #2 is approximately 33.33%.