Assume there are two urns with 4 balls each. Each ball has a certain probability to be picked. For example:
Urn1:
- Ball A: 0.7
- Ball B: 0.1
- Ball C: 0.1
- Ball D: 0.1
Urn2:
- Ball E: 0.3
- Ball F: 0.2
- Ball G: 0.25
- Ball H: 0.25
Notice the numbers next to the balls indicate the pick probability (e.g. there's a chance of 0.7 to pick Ball A when picking from Urn1).
My question is then: When 5 balls from each urn are picked, what's the probability for the event
Z=(A and E were picked at least once OR C and G were picked at least once)
i.e. what's the probability for P(Z)?
To clarify what I mean here are some example sequences:
picked from Urn1 = {A, A, C, D, A}
picked from Urn2 = {H, H, F, H, F}
A was picked from Urn1 but E was not from Urn2. C was picked from Urn1 but G was not from Urn2. Therefore the event Z did not occur.
picked from Urn1 = {A, A, A, D, A}
picked from Urn2 = {E, G, H, G, H}
A was picked from Urn1 and E from Urn2. Therefore the event Z occurred.
picked from Urn1 = {A, C, A, D, A}
picked from Urn2 = {E, G, H, G, H}
A was picked from Urn1 and E from Urn2. Additionally, C was picked from Urn1 and G from Urn2. Therefore the event Z occurred.
picked from Urn1 = {B, C, D, B, C}
picked from Urn2 = {G, G, H, H, H}
A was not picked from Urn1. However, C was picked from Urn1 and G from Urn2. Therefore the event Z occurred.
Let $P(A)$ represent the probability that $A$ will be picked among the five balls from the first urn; similarly for $P(C), P(E), P(G)$.
And let $P(A\wedge E)$ represent the probability that $A$ and $E$ are both chosen, $P(A\vee C)$ represent the probability that either $A$ or $C$ was chosen, and similarly for letters other than $A$ and $E$ or $C$.
What we want is $$P((A\wedge E) \vee (C\wedge G))$$ This is almost the same as $P(A\wedge E) + P (C\wedge G)$ but one would have double-counted cases where all of $A,C,E,G$ were chosen. Thus $$P((A\wedge E) \vee (C\wedge G)) = P(A\wedge E) + P (C\wedge G) - P(A\wedge C \wedge E\wedge G)$$
Now we can use the fact that the two urns happenings are completely independent to write this as $$P((A\wedge E) \vee (C\wedge G)) = P(A) \cdot P(E) + P(C) \cdot P(G) - P(A\wedge C) \cdot P(E \wedge G)$$
$P(A)$ is easy: The probability of $A$ not being chosen is $(1-0.7)^5$ so $P(A) = 1 - (0.3)^5 = 0.99757$
Similarly $$P(E) = 1-(1-0.7)^5=0.83193 \\ P(C) = 1-(1-0.9)^5=0.40951\\P(G) = 1-(1-0.75)^5=0.7626953125 $$
We still have not found $P(A\wedge C)$ or $P(E \wedge G)$ and these are not a simple matter of multiplying individual probabilities, because the probability of getting a $C$ is different if we know that at least one pick was an $A$. We can get $P(A\wedge C)$ by the trick based on calculating $P(A \vee C)$ as follows: $$ P(A \vee C) = 1 - P(\lnot(A \vee C)) = 1 - (1-[0.7+0.1])^5 = 1-(0.2)^5 = 0.99968 \\ P(A \vee C) = P(A) + P(C) - P(A \wedge C) \\ P(A \wedge C) = P(A) + P(C) - P(A \vee C) = 0.4074 $$ and $$ P(E \vee G) = 1 - P(\lnot(E \vee G) = 1 - (1-[0.3+0.25])^5 = 1-(0.45)^5 \approx 0.981547 \\ P(E \wedge G) = P(E) + P(G) - P(E \vee G) \approx 0.61307804074 $$ Putting it all together we have $$ P((A\wedge E) \vee (C\wedge G)) = P(A) \cdot P(E) + P(C) \cdot P(G) - P(A\wedge C) \cdot P(E \wedge G) \\ \approx 0.892472 $$