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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:

  1. to obtain a green ball from $A$ AND from $B$
  2. to obtain a green ball only from $A$ OR only from $B$
  3. to obtain a black ball from $A$ OR red from $B$
  4. 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 ?

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1 Answer 1

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(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. $$1-\frac25\frac15=\frac{23}{25}$$

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  • $\begingroup$ Thank you very clear $\endgroup$
    – sound wave
    May 2, 2018 at 10:45

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