An unfair coin with Pr[H] = 2/3 is flipped. If the flip results in a head, then a marble is selected from an urn containing 9 red, 1 white, and 1 blue marbles. If the flip results in a tail then a marble is selected from an urn containing 10 red and 3 white marbles. If the marble selected is white, then what is the probability that a flip resulted in a head?
To find the probability that a flip resulted in a head given that a white marble was selected, we can use Bayes' theorem.
Let's denote the event "A" as the event of flipping a head, and the event "B" as the event of selecting a white marble. We want to find the probability of "A" given "B", denoted as P(A|B).
According to Bayes' theorem:
P(A|B) = (P(B|A) * P(A)) / P(B)
P(B|A) is the probability of selecting a white marble given that a head was flipped. This is given as 1 white marble out of a total of 9 red + 1 white + 1 blue marbles, which is 1/11.
P(A) is the probability of flipping a head, which is given as 2/3.
P(B) is the probability of selecting a white marble, which can be calculated by considering the two cases: flipping a head and selecting a white marble, or flipping a tail and selecting a white marble. We need to find the probability of each case and sum them.
Case 1: Flipping a head and selecting a white marble.
P(H and W) = P(H) * P(W|H) = (2/3) * (1/11)
Case 2: Flipping a tail and selecting a white marble.
P(T and W) = P(T) * P(W|T) = (1/3) * (3/13)
P(B) = P(H and W) + P(T and W)
Now we can substitute all the values into Bayes' theorem:
P(A|B) = (P(B|A) * P(A)) / P(B)
= [(1/11) * (2/3)] / [(2/3) * (1/11) + (1/3) * (3/13)]
Simplifying the expression gives:
P(A|B) = (2/11) / [(2/11) + (1/13)]
Calculating further:
P(A|B) = (2/11) / (26/143 + 11/143)
= (2/11) / (37/143)
= (2/11) * (143/37)
≈ 0.857
Therefore, the probability that a flip resulted in a head given that a white marble was selected is approximately 0.857 or about 85.7%.