Why if I use Bayes' formula I'm right and if I use conditional probability formula I'm not?

Question

Let $\Omega$ a discrete probability space we have $U_1$ and its opposite $U_2$ :

$P(U_1) = 0,7$ and $ P(U_2) = 0,3$

$P(A|U_1) = 0,05$ and $P(A|U_2) = 0,06$ What's $P(U_2|A)$?

My solution using conditional probability :

$P(A) = P(A|U_1)\cdot P(U_1) + P(|U_2) \cdot P(U_2) = 0,05 \cdot0,7 + 0,06 \cdot 0,3 = 0,368 $

$P(U_2|A) = P(U_2\cap A)/P(A) = 0,018/0,368 = 0,048$

That's incorrect, the professor uses Bayes' formula and he gets $18/53$

Why is my solution wrong?

Answer 1

$$P(U_2\mid A)=\frac{P(U_2\cap A)}{P(A)}.$$

$$P(A)=P(A\mid U_1)P(U_1)+P(A\mid U_2)P(U_2)=\frac5{100}\frac7{10}+\frac6{100}\frac3{10}=\frac{53}{1000}.$$

$$P(U_2\cap A)=P(A\mid U_2)P(U_2)=\frac6{100}\frac3{10}=\frac{18}{1000}.$$

$$P(U_2\mid A)=\frac{\frac{18}{1000}}{\frac{53}{1000}}=\frac{18}{53}.$$

I don't know why the Bayes formula had to be used, but based on the definitions and the data given the right result is given above.

Answer 2

By Bayes' Rule, and Law of Total Probability is the same method as you used: $$\begin{split}\mathsf P(U_2\mid A) & =\dfrac{\mathsf P(A\mid U_2)\mathsf P(U_2)}{\mathsf P(A)}\\ &= \dfrac{\mathsf P(A\mid U_2)\mathsf P(U_2)}{\mathsf P(A\mid U_1)\mathsf P(U_1)+\mathsf P(A\mid U_2)\mathsf P(U_2)}\\ &= \dfrac{0.06\cdot 0.3}{0.05\cdot 0.7+0.06\cdot 0.3}\\ &= \dfrac{18}{53}\end{split}$$

So it's just a matter of correct calculations.