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I'm having a little trouble understanding this problem, and am wondering what I'm missing..

*A jar contains blue and red marbles, and you draw two marbles consecutively without replacement. Here are known probabilities:

-Probability of selecting blue then red marble is 30%

-Probability of selecting red marble on the first draw is 50%

What's the probability of selecting a blue marble, after selecting a red one?

It seems like by conditional probability, $$P(blue|red)=\frac{P(blue)*P(red|blue)}{P(red)}$$ and by substituting the known probabilities (by deducing that $P(blue)=.50$), get that $P(blue|red)=0.3$

but then I also know $P(red \cup blue) = P(red) + P(blue) - P(red \cap blue) $, so if these events are mutually exclusive, then wouldn't that mean $P(red \cap blue) = P(blue) * P(red|blue) = 0$?

How does this make sense? Are there any assumptions here that can't be made?

Thank you

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You're assigning the event $blue$ to "draw blue on first try" and "draw blue on second try", but they are not equivalent events. So your first equation doesn't hold. $P(blue)$, which you're calling the probability that we draw blue first, doesn't belong in the same equation as $P(blue|red)$, which you're calling the probability of drawing blue second after drawing red.

It helps to consider a concrete number of marbles. Let $b$ and $r$ represent the number of red and blue marbles we start with. We know $b=r$. Let's break down the statement "probability of selecting blue then red marble is 30%".

$0.3=P(blue)P(redafterblue)=0.5*P(redafterblue)$

Which means that $\frac{r}{r+b-1}=\frac{3}{5}$.

Solve for $b$ and $r$ by plugging in $\frac{b}{2b-1}=\frac{3}{5}$ and getting $b=3$. Now it should be easy to get P(select blue after red).

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  • $\begingroup$ Thank you so much! I understand your approach, but it doesn't look like it uses Bayes rule-- am I right to say that? Just want to double check $\endgroup$ – Stats_anon Jan 7 '18 at 23:07
  • $\begingroup$ It does use Bayes' rule. 0.3=P(red on 2nd draw and blue on first draw)=P(blue on first draw)*P(red on 2nd draw|blue on first). Divide both sides by P(blue on first) and you've got Bayes' in the form that you gave in the question. $\endgroup$ – Anshu says Reinstate Monica Jan 8 '18 at 15:03

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