# Drawing marbles out of a bag…

The infamous drawing marbles from a bag type question that I am used to seeing has became more complicated. Given a bag with 2 green marbles, 5 red marbles, and 3 yellow marbles. I then draw 2 marbles out without replacing them, what is the probability that there is at least one red marble?

I have worked out 7/9 as my answer on paper, but I would like to know the correct approach to solving this problem. Perhaps in a more abstract, but understandable form?

-
Probability is $0$, as you cannot draw a ball from a bag containing marbles! :-) –  user21436 Mar 1 '12 at 2:44
Ha! I fixed the typo. Though I would argue that a marble is a ball... –  Austin Henley Mar 1 '12 at 2:47

The easiest approach is to calculate the probability that you get no red marble and subtract that from $1$.

In order to get no red marble, you must (a) get a non-red marble with your first draw and then (b) get another non-red marble on your second draw. The probability of (a) is $\frac5{10}=\frac12$, since there are $2+3=5$ non-red marbles. Similarly, if (a) occurred, the probability of (b) is $\frac49$, since there are now only nine marbles in the bag, of which four are non-red. The combined probability of (a) and (b) occurring is therefore $$\frac12\cdot\frac49=\frac29\;,$$ and the probability of getting at least one red marble is $$1-\frac29=\frac79\;,$$ just as you calculated.

Of course you could work the problem directly, by calculating the probability of getting exactly one red marble and the probability of getting two red marbles and adding these probabilities. If you do it directly, you have $$\frac12\cdot\frac49=\frac29$$ as the probability of getting two red marbles; $$\frac12\cdot\frac59=\frac5{18}$$ as the probability of getting a red followed by a non-red marble; and $$\frac12\cdot\frac59=\frac5{18}$$ as the probability of getting a non-red followed by a red marble, for a total probability of $$\frac29+\frac5{18}+\frac5{18}=\frac79$$ of getting at least one red marble.

Clearly, though, it’s easier to consider the one ‘bad’ case of getting no red marbles than to have to work with the several ‘good’ cases. Had you been drawing three marbles instead of four, and had the problem asked for the probability of getting at least two red marbles, it would have been just about a toss-up, since there are two cases either way: $0$ or $1$ red marble (‘bad’) versus $2$ or $3$ red marbles (‘good’).

-
Brian I think you are on this website far too much, you keep answering my questions! But thank you, I am taking my first graduate level math class and it likes to make everything complicated. –  Austin Henley Mar 1 '12 at 2:49
@Austin: I’m retired, so I can afford the time; I still enjoy both mathematics and teaching, so it makes a nice hobby. And you’re very welcome. –  Brian M. Scott Mar 1 '12 at 2:59

The probability there is at least one red is the probability there is one red plus the probability there is two reds.

So

$$\text{Pr}(\text{2 reds}) = \frac{5}{10} \cdot \frac{4}{9} = \frac{2}{9}$$

To draw 1 red we can draw it on the first ball or on the second.

$$\text{Pr}(\text{1 red}) = \frac{5}{10}\cdot\frac{5}{9} + \frac{5}{10} \cdot \frac{5}{9} = \frac{5}{9}$$

Thus $$\text{Pr}(\text{At least 1 red}) = \frac{2}{9}+\frac{5}{9} = \frac{7}{9}$$

as you claim.

-

You can solve using combination as follows:

Probability at least one red means $p(\text{at least 1 red}) = 1-p(\text{not red})$

$1-\frac{_5C_2}{_{10}C_2}$ because $_{10}C_2$ is the total out come=45 and $_5C_2 =10$ so

$$p(\text{at least one red})= 1-0.22222=0.777777$$

-
Welcome to MSE. It helps if you format using MathJax (see FAQ). This answer could use a bit of cleanup. Regards –  Amzoti Aug 1 '13 at 4:25
Cleaned it up so that it's at least a bit more legible, but it could still use some work, mainly in explaining what was done a bit better. –  Ataraxia Aug 1 '13 at 4:52