# Having trouble applying Bayes Theorem to problem

For a class I am intending to use Bayes Theorem to solve a problem. The matter is that there are much easier ways to solve the problem and I don't know how to find the quantities to apply Bayes Theorem.

Here is the problem:

The blood type distribution in the United States is type A, 41%; type B, 9%; type AB, 4%; and type O, 46%. It is estimated that during World War II, several errors were made in typing blood. In fact, 4% of inductees with type O blood were listed as having type A; 88% of those with type A were correctly listed; 4% with type B blood were listed as type A; and 10% with type AB blood were listed as type A. A soldier was wounded and brought to surgery. He was listed as having type A blood. What is the probability that this is his true blood type?

To solve the problem, I multiplied the falsely listed blood type groups and added them, then divided by total probability:

(0.41 x 0.88) (A listed as A)+    (0.9 x 0.04) (B listed as A) + (0.04 x 0.1) (AB listed as A) + (0.46 x 0.04) (O listed as A) = 0.3868

A listed as A is 0.3608, so  0.3608 / 0.3868 = 0.933.

0.933 is the probability.


But what of Bayes' Theorem for the solution? I came up with this much more convoluted response:

 Let D denote that you are type A, let C denote that you are correctly listed:

P(D|C) is desired.

P(D|C) = P(C|D) * P(D)/ (P(C|D) * P(D) + P(C|D') * P(D'))

P(C|D) = P(D&C)/P(D) = 0.41 * 0.88 / 0.41 = 0.88
P(D) = 0.41

P(C|D') = 0.59 * 0.88 / 0.59 = 0.88
P(D') = 0.59

So, P(D|C) = 0.88 * 0.41 / ((0.88 * 0.41) + (0.88 * 0.59)) = 0.41

0.41 is the probability.


Obviously very different than my other answer.... and I am confident is my first answer vs. the Bayes' Theorem output.

What am I doing that make my Bayes' Theorem output wrong? Why can't I correctly apply the theorem?

-

$$P(A|L)=\frac{P(L|A)P(A)}{P(L|A)P(A)+P(L|B)P(B)+P(L|AB)P(AB)+P(L|O)P(O)}$$