I have the following problem: I have 2 urns, one which contains which contains 20% red balls and another which contains 80% red balls. Each time I select (with replacement) a ball, I am blind and have an equal chance of selecting from each urn.

What is the probability of selecting a red ball?

It seems to me in the long run I should obtanin 50% red balls so it's 0.5. But this does seem misleading somehow?

Many thanks!

  • 2
    $\begingroup$ This looks like a homework question to me. If this is true, could you please add the self-study tag? Thank you. Why do you think your result is misleading? $\endgroup$ Commented Sep 30, 2013 at 18:25

2 Answers 2

                                 /--- 80% --- Red
                           /     \--- 20% --- Carmine
-_- (you blindfolded)
                           \     /--- 20% --- Red
                                 \--- 80% --- Amaranth

You've got a $50\%$ chance of going up. At that point you have an $80\%$ chance of scoring Red. Good thing you're blindfolded, otherwise you might mistake Carmine for Red. Anyways, what's the probability of you following this path and hitting red?

$$ P(\text{up and red}) = 0.5 \cdot 0.8 = 0.4 $$

What about going down? Well, since you're not given the chance of mistaking Amaranth for Red (I really hope no one knows what Amaranth looks like), the math is pretty much the same.

$$ P(\text{down and red}) = 0.5 \cdot 0.2 = 0.1 $$

Since you're equally like to go both ways your probabilities add up. Thus you end up with

$$ P(\text{red}) = 0.4 + 0.1 = 0.5 $$

Save for the confusing colors and color names this should not seem too misleading. Good luck playing the game. I hope they give you good odds.


I find it quite useful to draw a tree diagram for such problems:

Tree diagram

The numbers represent probabilities. You have select each urn with probability $0.5$. In the first urn, there are $20\%$ red balls, so the probability to draw a red ball is $0.2$. For the second urn, the probability to draw a red ball is $0.8$. The red probabilities at the end of the branches are the products of probabilities leading to these branches. The probabilities at the end of the tree have to sum to $1$. I have marked the outcomes were you draw a red ball with a blue frame. To combine probabilities across the branches of the tree, add them up. From the tree diagram it becomes obvious that your result is correct: $P(\text{Red ball})=0.1 + 0.4 = 0.5$.


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