Probability Question

Question

A specially made pair of dice has only one- and two-spots on the faces. One of the dice has three faces with a one-spot and three faces with a two-spot. The other die has two faces with a one-spot and four faces with a two-spot. One of the dice is selected at random and then rolled six times. If a one-spot shows up four times, what is the probability that it is the die with the three one-spots?

Answer 1

Rick:

 

Application of Bayes Theorem.

 

Identify the towo dice as two conditions called C1 and C2. We are given that one die is selected at random. We are also given the outcome, O, say, that one-spot has shown up four times in six rolls.  Bayes Theorem gives you the probabilities of the conditions P(C1|O) and P(c2|O) given the outcome o.  Consider P(C1|O) first.

 

The formula is based on the joint probability written two ways: P(O and C1) = P(O|C1)P(C1) (one way) and

 

P(O and C1) = P(C1 | O) P(O)  (another way).  Equating the two,

 

P(C!|O) = P(O|C1)P(C1) / P(O), provided P(O) is not zero.

 

Now by the law of total probability, P(O) = P(O|C1)P(C1) + P(O|C2)P(C2). Hence

 

P(C!|O) = P(O|C1)P(C1) / [ P(O|C1)P(C1) + P(O|C2)P(C2)], provided the denominator is not zero.

 

Now we are ready to compute the quanties and answer the question.

 

P(O|C1) = (0.5)⁴(0.5)² = (0.5)⁶.  P(O|C2) = (1/3)⁴(2/3)².  P(C1) = P(C2) = 0.5 (die chosen at random).  Hence

 

P(C!|O) = P(O|C1)P(C1) / [P(O|C1)P(C1) + P(O|C2)P(C2)]

 

            = (0.5)⁶ (0.5) /[ (0.5)⁶(0.5) + (1/3)⁴(2/3)²(0.5)]

 

See that (0.5) cancels.  I will leave the rest of the arithmetic to you. Post a comment if you have nay questions.

 

Dr. G.