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An urn contains $10$ red marbles and $10$ black marbles while a second urn contains $25$ red marbles and an unknown number of black marbles. A random marble will be selected from each urn and the probability that both marbles are the same will be determined. A hint was given by the teacher: the probability does NOT depend on the number of unknown marbles. Verify that this is the case.

Let's call $N$ the unknown number of marbles.

I wrote out all possible ways to select a marble from each urn, selecting a red marble from both urn $1$ and urn $2$, and selecting a black marble from urn $1$ and $2$ and this is what I got:

  • Number of ways to select a marble from each urn: $ \binom{20}{1}\binom{25+N}{1}$

  • Number of ways to select $1$ red marble from both urn $1$ and urn $2$: $\binom{10}{1}\binom{25}{ 1}$

  • Number of ways to select $1$ black marble from urn $1$ and urn $2$: $\binom{10}{1}\binom{N}{ 1}$

And this is what I got as my final equation to finding out the probability of selecting the same color marble from each urn: $\dfrac{\binom{10}{1}\binom{25}{ 1}+\binom{10}{1}\binom{N}{ 1}}{\binom{20}{1}\binom{25+N}{1}} $

I am confused on how the probability doesn't depend on the unknown number of black marbles in urn $2$? Any help would be much appreciated, thank you so much!

PS: I also searched through stack exchange for a problem similar to this and I couldn't find one. If this question was asked already, then I apologize!

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4 Answers 4

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A short explanation of why this is, is that instead of picking both marbles simultaneously, let us pick a marble from the strange bag first.

Once having done so, it will be some color, either black or red. Now, pick a marble from the one with an even amount of each color. Regardless which color was chosen in the first step, the probability that you pick the same color will be $\frac{1}{2}$.

Worded with symbols, let $B_1,R_1,B_2,R_2$ represent the events of picking a black marble or red marble from the first or second bag respectively. Let the "second bag" be the one with the unknown number of black marbles.

We have:

$$\begin{array}{rl}Pr(\text{colors are same}) &= Pr((B_2\cap B_1)\cup (R_2\cap R_1)) \\&= Pr(B_2)Pr(B_1|B_2)+Pr(R_2)Pr(R_1|R_2)\\ &=\frac{1}{2}Pr(B_2)+\frac{1}{2}Pr(R_2)\\ &=\frac{1}{2}(Pr(B_2)+Pr(R_2))\\ &=\frac{1}{2}\end{array}$$

This uses the multiplication principle that $Pr(A\cap B) = Pr(A)Pr(B|A)$

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    $\begingroup$ wow great explanation from everyone. Thank you! Makes a LOT more sense now! $\endgroup$
    – King
    Dec 14, 2015 at 1:20
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Suppose the probability of drawing a red marble from the second urn is $p$. It would then follow that the probability of drawing two red marbles is $$\frac 12p$$

But then the probability of drawing a black marbles from the second urn is $1-p$, whence the probability of drawing two black marbles is $$\frac 12(1-p)$$

As these are disjoint events we add the probabilities to see that the probability that the two random marbles share a color is $$\frac 12(p+(1-p))=\frac 12$$

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  • $\begingroup$ This is the cleanest solution. $\endgroup$
    – user217285
    Dec 14, 2015 at 1:44
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Hint:

$\dfrac{\binom{10}{1}\binom{25}{ 1}+\binom{10}{1}\binom{N}{ 1}}{\binom{20}{1}\binom{25+N}{1}} =\frac{\binom{10}{1}\cdot\left[\binom{25}{ 1}+\binom{N}{ 1} \right]}{20\cdot(25+N)}=\frac{10\cdot (25+N)}{20\cdot (25+N)}$

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Let $p$ be the probability of drawing a red marble from the second urn, so $1-p$ is the probability of drawing a black marble. Then the probability that the marbles match is

$$ P(\text{match}) = \left(\frac{1}{2}\right)p + \left(\frac{1}{2}\right)(1-p) = \frac{1}{2} $$

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