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

Answer:

Given,

The table is given as follows i.e.,

x	0	1	2	3	4
P(x)	1/5	1/5	1/5	1/5	1/5

P(getting sequence of blue,green,blue) = P(1) + P(2) + P(3)

= P(one blue marble)*P(drawing blue marble or ball)

substitute the values

= 1/5(1/4 * 3/4 * 1/4) + 1/5(2/4 * 2/4 * 2/4) + 1/5(3/4 * 1/4 * 3/4)

= 0.0094 + 0.025 + 0.0281

= 0.0625

Let us consider,

Baye's theorem

P(A|B) = P(B|A)*P(A) / P(B)

= P(getting BGB|X=1)*P(X = 1)/P(getiig BGB)

= [ (1/4*3/4*1/4)*(1/5) ] / 0.0625

= 0.009375/0.0625

= 0.15

= 15%

Required probability = 0.15