Probabilities, with solutions(bayes' theory)

Question

I've been doing some advanced probabilities and would like to check up on answers and solutions. Any help and corrections will be appreciated

1. One urn has 16 White balls and 6 black balls, second one has 31 white balls and 30 black ones. a ball is taken out of randomly selected urn. it is white. What is the probability of taking another white ball out of the same urn.

first urn 16W, 6B balls. Second urn 31W, 30B balls

to take a white ball out of first urn the probability is $\frac{16}{22}$, to take a white ball out of second urn probability is $\frac{31}{61}$

and if we take second ball from the same urn as before the answer should be $\frac{16}{22}\ast\frac{15}{21}+\frac{31}{61}\ast\frac{30}{60}$

2. The patient is sick with one of the following disease C1, C2, C3. it is known that C2 and C3 are twice as common as C1. the disease is diagnosed with success rate of C1-0.87, C2-0.51, C3-0.51. The patients test was successful. What is the probability that disease is in a form of C1?

to get C1 the chance is 1 in 5 as C2 and C3 are twice as likely to occur. we also know that the test was successful, but we need to make sure it was only successful on C1 and not on other variants

test was successful and it was C1 should be $\frac15\cdot0.87$

test was successful and it was not C2 should be $\frac35\cdot0.51$

test was successful and it was not C3 should be $\frac35\cdot0.51$ , we combine them and we get

$\frac15\cdot0.87+\frac35\cdot0.51+\frac35\cdot0.51=0.786$

there is one more exercise that i was not able to solve and just in case anyone is interested in helping me resolve it i will link it balls and probabilities(bayes' theory)

My sincere apologies for my lackluster English.

Answer 1

Actually in the first case, you must also consider the probability of choosing urn 1 or urn 2. Assuming $P(U_1)=P(U_2)=\dfrac{1}{2}$, the answer in the first case must be $$\dfrac{1}{2}\cdot\dfrac{16}{21}\cdot\dfrac{15}{20}+\dfrac{1}{2}\cdot\dfrac{31}{61}\cdot\dfrac{30}{60}$$

In the second case, let us assume $C_1$ be the event of having C1 disease, $C_2$ be the event of having C2 disease, $C_3$ be the event of having C3 disease and $S$ be the event of successful test.

Now as per Bayes theorem, $$P(C_1|S)=\dfrac{P(C_1)\cdot P(S|C_1)}{P(S)}$$

But $P(S)=P(C_1)\cdot P(S|C_1)+P(C_2)\cdot P(S|C_2)+P(C_3)\cdot P(S|C_3)$ $$P(S)=\dfrac{1}{5}\cdot 0.87+\dfrac{2}{5}\cdot 0.51+\dfrac{2}{5}\cdot 0.51=\dfrac{2.91}{5}$$ $$\Rightarrow P(C_1|S)=\dfrac{\Big(\dfrac{1}{5}\Big) 0.87}{\dfrac{2.91}{5}}=0.298$$