Question

A classic experiment in the 1960s eloquently illustrated the technical side of conservatism bias. The researchers presented subjects with two urns: one containing 3 blue balls and 7 red balls, the other containing 7 blue balls and 3 red ones. Subjects were given thisinformation and then told that someone had drawn randomly 12 timesfrom one of the urns [with replacement]. Subjects were told that thisdraw yielded 8 reds and 4 blues. They were then asked, "What is theprobability that the draw was made from the rst urn?" Formulate the problem in terms of Bayes' Rule. State all of your assumptions and all assignments explicitly.

Answer 1

Answer:

Step-by-step explanation:

I urn: 3 blue 7 red

II urn: 7 blue and 3 red

When draw is done with replacement no of red or blue will be binomial with equal probability for each trial

Let B = draw yielded 8 reds and 4 blues.

A1= Drawn from I urn

A2 = Drawn from II urn

A1 and A2 are mutually exclusive and exhaustive

Required probability = P(A1/B) = $\frac{P(A1B)}{P(A1B)+P(A2B)}$

P(A1B) = $(\frac{7}{10} )^8*(\frac{3}{10} )^4\\=\frac{7^8(3^4)}{10^{12} }$

P(A2B) = $(\frac{3}{10} )^8(\frac{7}{10} )^4\\=\frac{3^8(7^4)}{10^{12} }$

P(A1/B) =$\frac{7^8(3^4)}{7^8(3^4+3^8(7^4)}$