We consider two urns that contain blue balls and red balls. Urn $1$ has $9$ blue balls and $1$ red one. Urn $2$ has $8$ blue balls and $2$ red ones. We now draw one ball from each urn.

(1) Find the probability of at least one ball being red.

(2) Find the probability of exactly one ball being red.

(3) You know know that exactly one of the balls drawn is red. Find the probability that the ball comes from Urn $1$.

(1) ($\dfrac{1}{10} \cdot \dfrac{2}{10}) + (\dfrac{1}{10} \cdot \dfrac{8}{10}) + (\dfrac{9}{10} \cdot \dfrac{2}{10}) = 0.28$.

(2) $(\dfrac{1}{10} \cdot \dfrac{8}{10}) + (\dfrac{9}{10} \cdot \dfrac{2}{10}) = 0.26.$

(3) Let $R$ be the event of drawing exactly one red ball ($P(R) = 0.26$) and $U_1$ the event of drawing from Urn $1$.

$$P ( U_1| R) = ?$$

How are we supposed to find $P(U_1)$? I tried to model the question using a (single) tree diagram but was unsuccessful.

  • 2
    $\begingroup$ I think you're thinking about Bayes' theorem when you say "how to find $P(U_1)$", but you don't need that here. Just the definition of conditional probability is needed.$$P(U_1|R)=\frac{P(U_1\cap R)}{P(R)}=\frac{\textrm{prob of picking red ball when urn 1 is chosen}}{\textrm{prob of picking exactly one red ball}}$$ $\endgroup$ – Prasun Biswas Oct 20 '20 at 21:19
  • $\begingroup$ @PrasunBiswas Indeed. This clears things up, thank you! $\endgroup$ – Katja Oct 20 '20 at 21:22
  1. $$\frac {1}{10}\times\frac{8}{10}+\frac {9}{10}\times\frac{2}{10}+\frac {1}{10}\times\frac{2}{10}=1-\frac {9}{10}\times\frac{8}{10}$$

  2. $$\frac {1}{10}\times\frac{8}{10}+\frac {9}{10}\times\frac{2}{10}$$

  3. $$P( U_1| R)=\frac {P(U_1\cap R)} {P(R)}= \frac{\frac {1}{10}\times\frac{8}{10}}{\frac {1}{10}\times\frac{8}{10}+\frac {9}{10}\times\frac{2}{10}}$$


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