Question

An urn contains 4 marbles, either blue or green. The number of blue marbles is equally likely to be 0, 1, 2, 3, or 4. Suppose we do 3 random draws with replacement, and the observed sequence is: blue, green, blue. What is the probability the urn contains just 1 blue marble?

Answer 1

B0 = Event of urn contains 0 blue marbles

B1 = Event of urn contains 1 blue marbles

B2 =  Event of urn contains 2 blue marbles

B3 = Event of urn contains 3 blue marbles

B4 = Event of urn contains 4 blue marbles

The number of blue marbles is equally likely to be 0, 1, 2, 3, or 4.

Therefore,

P(B0)=P(B1)=P(B2)=P(B3)=P(B4) = 1/5 =0.2

X : Event of drawing blue, green, blue in sequence  with replacement

P(X|B0) = Probability drawing blue, green, blue in sequence  with replacement = 0 (as there are no blue marbels in the urn)

P(X|B1) : (1/4)x(3/4)x(1/4) = 3/64 (As there are 1 blue marble and 3 green marbles; Probability of drawing a blue marble each time is 1/4 and green marbles is 3/4 as the draw with replacement)

Similary

P(X|B2) = (2/4)x(2/4)x(2/4) = 8/64

P(X|B3) = (3/4)x(1/4)x(3/4) = 9/64

P(X|B4) = 0 (as when the urn contains all blue marbles, number of green marbles = 0, so we can not draw green marble in the second draw)

Suppose we do 3 random draws with replacement, and the observed sequence is: blue, green, blue,  probability the urn contains just 1 blue marble = P(B1|X)

By Bayes theorem,

P(B1)P(X|B1) = (1/5) x (3/64) = 3/320

P(B0)P(X|B0) = 0

P(B2)P(X|B2) = (1/5) x (8/64) = 8/320

P(B3)P(X|B3) = (1/5) x (9/64) = 9/320

P(B4)P(X|B4) = 0

P(B0)P(X|B0) + P(B1)P(X|B1) + P(B2)P(X|B2) + P(B3)P(X|B3) + P(B4)P(X|B4) = 0+3/320+8/320+9/320+0 = 20/320

Suppose we do 3 random draws with replacement, and the observed sequence is: blue, green, blue,  probability the urn contains just 1 blue marble = 3/20 = 0.15