Bayesian Statistics (The Cookie Problem) (3)

The cookie problem:

” Suppose there are two full bowls of cookies. Bowl 1 has 10 sugars and 30 mints, while Bowl 2 has 20 of each. Fred picks a bowl at random, and then a random cookie. What is the probability that if Fred picks a mint cookie, it came from Bowl 1? “

H1=cookie from Bowl 1

H2=cookie from Bowl 2

P(H1)=P(H2)=1/2 (because picking a bowl is equal, thus 50/50)

P(E|H1)= likelyhood of getting a mint cookie from bowl 1 = 30/40=3/4

P(E|H2)=likelyhood of getting a mint cookie from bowl 2=20/40=1/2

Logically, we can make an argument that because Bowl 1 has more mint cookies, there is a greater likelyhood that it came from Bowl 1

Let’s plug values in!

P(H1|E)=P(H1)P(E|H1) / P(E)

= (1/2)(3/4)/(P(E))

P(E)=what is the total possibility of a mint? =50/80=5/8

= (1/2)(3/4)/(5/8)=3/5=60%