Marbles Probability, Should Be Easy

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Jun 7th 2011, 10:48 AM
ragnar

Marbles Probability, Should Be Easy

I'm looking at this question, and I know the answer. You have 10 blue and 12 red marbles, select two at random, what's the probability that you get at least one blue? I know I can find the probability of obtaining all red marbles, $\displaystyle \frac{12\cdot11}{22\cdot 21}$ and then find one minus this to get 5/7.

But I initially tried doing this by finding total number of ways of obtaining one blue marble (10) plus total number of ways of obtaining two blue marbles (10C2) divided by total number of ways of selecting any two marbles (22C2), and my answer is off by a factor of 3. Can anyone explain what's going wrong in this calculation?
Jun 7th 2011, 11:15 AM
Plato

Quote:

Originally Posted by ragnar

I'm looking at this question, and I know the answer. You have 10 blue and 12 red marbles, select two at random, what's the probability that you get at least one blue? I know I can find the probability of obtaining all red marbles, $\displaystyle \frac{12\cdot11}{22\cdot 21}$ and then find one minus this to get 5/7.

Try $\displaystyle \frac{\binom{10}{1}\binom{12}{1}}{\binom{22}{2}}$$\displaystyle +\frac{\binom{10}{2}}{\binom{22}{2}}$.
Jun 7th 2011, 02:30 PM
Soroban

Hello, ragnar!

Quote:

You have 10 blue and 12 red marbles, select two at random.
What's the probability that you get at least one blue?

I know I can find the probability of obtaining all red marbles, $\displaystyle \frac{12\cdot11}{22\cdot 21}$
and then find one minus this to get 5/7.

But I initially tried doing this by finding
total number of ways of obtaining one blue marble (10) . ← Here!
plus total number of ways of obtaining two blue marbles (10C2)
divided by total number of ways of selecting any two marbles (22C2),
and my answer is off by a factor of 3.
Can anyone explain what's going wrong in this calculation?

Number of ways to get one blue marble and one red marble:
. . $\displaystyle (_{10}C_1)(_{12}C_1}) \:=\:(10)(12) \:=\:120$

Number of ways to get two blue marbles:
. . $\displaystyle _{10}C_2 \:=\:45$

Number of ways to get at least one blue marble: .$\displaystyle 120 + 45 \:=\:165$

. . $\displaystyle \text{Therefore: }\;P(\text{at least one blue}) \:=\:\frac{165}{231} \;=\;\frac{5}{7}$