Probability-Bayes Theorem

• May 5th 2010, 10:23 PM
eri123
Probability-Bayes Theorem
Problem:

The Belgian 20-Frank coin (B20), the Italian 500-Lire coin (L500), and the Hong Kong 5-Dollar (HK5) are approximately the same size. Coin purse one (C1) contains six of each of these coins. Coin purse two (C2) contains nine B20s, six I500s, and three HK5s. A fair four-sided die is rolled. If the outcome is {1}, a coin is selected randomly from C1. If the outcomes belong to {2, 3, 4}, a coin is selected randomly from C2.
a) Find P(HK5).

b) Find P(C2|HK5).

I've drawn a tree diagram.
Now for a), is this correct? \$\displaystyle (3/4)(1/6)+(1/4)(1/3)=5/24 \$

And for b), do I use Bayes Theorem? Is this correct? \$\displaystyle ((3/4)(1/6))/((3/4)(1/6)+(1/4)(1/3))=3/5 \$

Thanks for the help!
• May 5th 2010, 11:13 PM
CaptainBlack
Quote:

Originally Posted by eri123
Problem:

The Belgian 20-Frank coin (B20), the Italian 500-Lire coin (L500), and the Hong Kong 5-Dollar (HK5) are approximately the same size. Coin purse one (C1) contains six of each of these coins. Coin purse two (C2) contains nine B20s, six I500s, and three HK5s. A fair four-sided die is rolled. If the outcome is {1}, a coin is selected randomly from C1. If the outcomes belong to {2, 3, 4}, a coin is selected randomly from C2.
a) Find P(HK5).

b) Find P(C2|HK5).

I've drawn a tree diagram.
Now for a), is this correct? \$\displaystyle (3/4)(1/6)+(1/4)(1/3)=5/24 \$

Yes.

Quote:

And for b), do I use Bayes Theorem? Is this correct? \$\displaystyle ((3/4)(1/6))/((3/4)(1/6)+(1/4)(1/3))=3/5 \$
You could use Bayes' theorem, but if you have a tree diagram I would just read the required sum off of that, but anyway your answer is correct.

CB