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There are $2$ black and $3$ yellow marbles in a bag. $2$ marbles are drawn randomly without replacement. What is the possibility that at least $1$ yellow marble is selected.

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    $\begingroup$ Hint: Find the probability of no yellow. $\endgroup$ May 18, 2013 at 6:57
  • $\begingroup$ I'm confused. Sorry, I can't comprehend probability that well :/ $\endgroup$
    – GreatEdgar
    May 18, 2013 at 7:10
  • $\begingroup$ Frankly I would suggest you consult your teacher. $\endgroup$
    – dfeuer
    May 18, 2013 at 7:13

1 Answer 1

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Imagine drawing the balls one at a time. The probability the first ball is black is $\frac{2}{5}$. Given that the first ball was black, the probability the second is black is $\frac{1}{4}$ (there is only $1$ black left, out of $4$ balls). So the probability both are black is $\frac{2}{5}\cdot\frac{1}{4}$.

It follows that the probability that there is at least one yellow is $1-\frac{2}{5}\cdot\frac{1}{4}$.

There are many other ways to solve the problem. Imagine the balls have ID numbers. There are $\binom{5}{2}$ equally likely ways to choose $2$ balls. There is only $1$ way to choose $2$ black, so the probability of $2$ black is $\frac{1}{\binom{5}{2}}$. It follows that the probability of at least one yellow is $1-\frac{1}{\binom{5}{2}}$.

In each of these solutions, we worked with the blacks, because it is easier. But we could work with the yellows. Either of the two ways we solved the problem can be adapted.

For example, we get at least one yellow if we draw BY (black, then yellow) or YB or YY.

The probability of BY is $\frac{2}{5}\cdot\frac{3}{4}$. In a similar way, find the probability of YB and of YY, and add up.

We can also do a counting argument. There are $\binom{5}{2}$ ways to choose $2$ balls. How many result in at least one yellow? We could have a yellow and a black, which can be chosen in $\binom{3}{1}\binom{2}{1}$ ways. Or we can have two yellow, which can be chosen in $\binom{3}{2}\binom{2}{0}$ ways. So our probability is $\frac{\binom{3}{1}\binom{2}{1}+ \binom{3}{2}\binom{2}{0}}{\binom{5}{2}}$.

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  • $\begingroup$ In the sentence that begins with "For example, ...", the author meant we get at least one yellow. $\endgroup$ Oct 30, 2018 at 20:39
  • $\begingroup$ @amWhy I do not understand what you are saying. Andre wrote "For example, we get $2$ yellow if we get BY (black, then yellow) or YB or YY." Clearly, he meant at least one yellow. $\endgroup$ Oct 30, 2018 at 21:08
  • $\begingroup$ "Imagine drawing the balls one at a time. The probability the first ball is black is $\frac{2}{5}$. Given that the first ball was black, the probability the second is black is $\frac{1}{4}$ (there is only $1$ black left, out of $4$ balls). So the probability both are black is $\frac{2}{5}\cdot\frac{1}{4}$. It follows that the probability that there is at least one yellow is $1-\frac{2}{5}\cdot\frac{1}{4}$." $\endgroup$
    – amWhy
    Oct 30, 2018 at 21:11

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