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There are $3$ bags, each containing $3$ coloured balls. The first two bags contain a red, green, and blue ball. The third bag contains $3$ red balls. You choose a bag at random. You then pull out a ball and replace it back into the bag. You then pull out another ball and determine that both of the balls you chose were red. What is the probability you chose the bag with the $3$ red balls?

P_redBag $= 1/3$ => P_redBall $= 1$
P_mixBag $= 2/3$ => P_redBall $= 1/3$

Bayes Rule:
$P(A|B) = P(A) * P(B|A)/P(B)$

$B = 1/3 * 1 + 2/3 * 1/9 = 0.4074$

The solution is $82$%

I cannot figure out how to derive that solution... I have tried:
$P(A|B) = 1/3 / (1/3 * 1 + 2/3 * 1/9)$

and several variations.

I assume that B is the probability of choosing a red ball, and A is the probability of choosing the bag full of red balls.

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    $\begingroup$ prob of two red balls from a mixed bag is 1/9 not 1/3 $\endgroup$ – almagest Sep 27 at 19:04
  • $\begingroup$ Thank you, I was wondering whether or not it should be 1/3 or 1/9. $\endgroup$ – Libor Zachoval Sep 27 at 19:15
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Let $M$ be the event of choosing a mixed bag (rgb), $R$ the event of drawing the red bag. So $P(M)=\frac23$, $P(R)=\frac13$, as only one of the three is all red, the other two are mixed.

$P(rr|R)=1$, obviously. $P(rr|M)=\frac{1}{9}$ by independence of the draws.

So $P(rr)=P(rr|M)P(M)+P(rr|M)P(M)=\frac13+ \frac23\frac19 = \frac{11}{27}$ Also $P(rr \land R)=P(R)P(rr|R)=\frac{1}{3}\cdot 1= \frac{1}{3}$

So $$P(R|rr)=\frac{P(R \land rr)}{P(rr)}= \frac{\frac{1}{3}}{\frac{11}{27}}=\frac{9}{11}$$ rounded to 82% if you're into that sort of thing.

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  • $\begingroup$ So my mistake was just not accounting for $P(rr|M) correctly. I might have actually tried that and the result $0.818$ might have thrown me off. Thank you for the full explanation. $\endgroup$ – Libor Zachoval Sep 29 at 9:36

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