Respuesta :
PART (a)
Let’s consider C as a random variable indicating the coin that we drew from a, b, c.
For network, C would be at the root and X1, X2 and X3 would be considered as children.
The Conditional Probability Table (CPT) for C:
C P (C)
a 1/3
b 1/3
c 1/3
The CPT for X1:
C X1 P (C)
a heads 0.2
b heads 0.6
c heads 0.8
PART (b)
According to the question, the following sequence is considered where P is the probability:
P(C|2 heads, 1 tails) = P (2 heads, 1 tails|C)P(C) / P(2 heads, 1 tails)
[tex]\alpha[/tex] P (2 heads, 1 tails|C)P(C)
[tex]\alpha[/tex] P (2 heads, 1 tails|C)
In the second line, we find that the constant of proportionality 1/P (2 heads, 1 tails) is independent of C, and P(C) is also independent of C in the last line, so, hypothetically, it is equal to 1/3.
In part a, we can see that X1, X2, and X3 are conditionally independent given C, therefore, for instance
P (X1 = tails, X2 = heads, X3 = heads | C = a)
= P (X1 = tails | C = a) P (X2 = heads | C = a) P (X3 = heads | C = a)
= 0.8 × 0.2 × 0.2 = 0.032
We would get the same probability for any order of 2 heads and 1 tails because the CPTs for each coin (a,b,c) are the same. Therefore, we have three such orders:
P(2 heads, 1 tails | C = a) = 3 × 0.032 = 0.096
P(2 heads, 1 tails | C = b) = 0.432
P(2 heads, 1 tails | C = c ) = 0.384
From the above results, it is concluded that coin b is most likely to have been drawn from the bag in the condition that is mentioned in the question.