There are 10 blue balls and 2 green balls inside a bag. Mr. Ali draws a ball inside the bag. If the drawn ball is blue, Mr. Ali will replace it with 4 blue balls inside the bag. If the drawn ball is green, Mr. Ali will replace it with 6 green balls inside the bag. After that, Ms. Bea draws a ball inside the bag. Let A be the event that Mr. Ali draws a green ball and B be the event that Ms. Bea draws a green ball. Determine the following probabilities.
- $P(B)$
I am confused if the probability of event B has two values or exactly one value. Since Mr. Ali will be the first one who will draw a ball in the given scenario, then there must be 2 possible values. Either$P_{A}(B)$ or $P_{\overline{A}}(B)$.
If $P(B)$ has only one value, I am thinking that it is the same as $P(A)$.
- $P_B(A)$ Probability of Event A given event B occured
This one is only possible to be solved using the formula for the conditional probability if $P(B)$ was already known. Since I am not sure how many values of $P(B)$ has, I am confused about how to get this value.
Any comments or suggestions will be much appreciated. Thank you in advance.
Evaluating the situation right before Bea's drawing:
(5/6) of the time, Bea will be faced with 13 blue, 2 green.
(1/6) of the time, Bea will be faced with 10 blue, 7 green.
Therefore
$$p(B) ~=~ \left[\frac{5}{6} \times \frac{2}{15}\right] ~+~ \left[\frac{1}{6} \times \frac{7}{17}\right].$$
I am interpreting the syntax of the second question to be chance of event B occurring assuming that event A has occurred. This will be
$$\frac{7}{17}.$$
Addendum
Apparently, I misinterpreted the second question. Instead, it is asking for
$$p(A|B) ~=~ \frac{p(A ~\text{and} ~B ~\text{both occuring})}{p(B)}.$$
Using the work already done at the start of this answer, that evaluates to
$$\frac{\frac{1}{6} \times \frac{7}{17}}{\left[\frac{5}{6} \times \frac{2}{15}\right] ~+~ \left[\frac{1}{6} \times \frac{7}{17}\right]}.$$