A urn $A$ contains $3$ green balls and $2$ black, a urn $B$ contains $4$ green and $1$ red. Each ball has the same probability to be chosen.
If we draw one ball from each urn, what are the probabilities:
- to obtain a green ball from $A$ AND from $B$
- to obtain a green ball only from $A$ OR only from $B$
- to obtain a black ball from $A$ OR red from $B$
- to obtain at least one green ball
This is what I got:
(1) $$P(1G \cap 2G) = P(1G) P(2G) = \frac{3}{5}\frac{4}{5} = \frac{12}{25}$$
(2) $$P(\text{"only 1G" } \cup \text{ "only 2G"}) = P(1G)P(\overline{2G}) + P(\overline{1G})P(2G) = \frac{3}{5}\frac{1}{5} + \frac{2}{5}\frac{4}{5} = \frac{11}{25}$$ (incompatible events)
(3) $$P(1B \cup 2R) = P(1B) + P(2R) - P(1B)P(2R) = \frac{2}{5} + \frac{1}{4} - \frac{2}{5}\frac{1}{4} = \frac{11}{20}$$ (compatible events)
(4) $$P(\text{"at least one green"}) = P(1G \cap 2G) + P(\text{"only 1G" } \cup \text{ "only 2G"}) = \frac{12}{25} + \frac{11}{25} = \frac{23}{25}$$
where $1G$ denotes the event "ball drawn from urn $A$ is green" and $\overline{1G}$ its negation.
Are these corrects ?
(1) and (2) are okay.
(3) should be $$\frac25+\frac15-\frac25\frac15=\frac{13}{25}$$
(4) is correct but there is a more efficient way: $1$ minus the probability that no green balls are drawn. $$1-\frac25\frac15=\frac{23}{25}$$