Bayes's Theorem

Question

Yesterday car lot Alpha had five Toyotas and one Chevrolet for sale. Car lot Beta had three Toyotas and five Chevrolets for sale. This morning Alan bought a car, choosing one of the two lots at random and then choosing a car at random from that lot. Leave your answers as fractions.

 

It can be shown that the probability that Alan bought a Toyota is 29/48. What is the probability Alan bought from Alpha given that he bought a Toyota? Give your answer both as a fraction and in decimal form rounded to two places. Note: In this situation, Bayes's theorem says
P(Alpha given Toyota) = P(Toyota given Alpha) × P(Alpha)/P(Toyota)

Answer 1

Make a tree, starting from one point at the left of a sheet of paper; make two branches, labeled alpha and beta, and place the probability 1/2 next to those labels on the 'branches.'

 

From the right end of the alpha branch, create two new branches, heading to the right; label the top one Chev and the other Toyota.  Alongside those labels place probabilities 1/6 and 5/6, respectively--reflecting the fact that lot Alpha has one Chev & 5 Toyota, total of 6 vehicles for sale.

 

From the right end of the beta branch, make two new branches, also heading to the right; label them Chev and Toyota, and give them probabilities 5/8 and 3/8, reflecting the fact that lot Beta has 5 Chevys & 3 Toyotas for sale.

 

You obtain four different probabilities at the end of each two-branch path, by multiplying the probabilities at each of the two particular branches.  That is, P(Chev & Alpha) = (1/2)(1/6) = 1/12.  The other probabilites are, from top to bottom, 5/12, 5/16 and 3/16.  A check shows that the sum of these four probabilities is 1, as it must be.

 

Now, apply the formula from Bayes Theorem.  Numerator is 5/12 times 1/2, i.e. P(Toyota|Alpha) times P(Alpha).

The denominator P(Toyota) is the sum of the probabilities of Toyota, from either lot, i.e. denom. = 5/12 + 3/16 = 29/48.

 

Final answer is then 10/29.