# What's the probability of randomly selecting 3 red pencils from a box containing 5 red, 3 blue, and 4 green pencils?

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I want someone to double check my answer

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Jake M. Share
Mar 7, 2018

1/22

#### Explanation:

There's a concept of replacement which makes a lot of these "balls in bags" questions a little tricky.

If we pick out a red pencil and then put it back in the bag and pick another, that's "with replacement". If we just pick three pencils at the same time, that's "without replacement".

In most cases, it makes more sense to be without replacement but that does make the math a little harder. It definitely seems like that's what you mean in this problem.

In that case, let's think about the probability of picking a red pencil the first time. The most important formula for all of statistics is
$\frac{\textrm{\nu m b e r o f w a y s \to \succ e e d}}{\textrm{\nu m b e r o f p o s s i b i l i t i e s}}$

Initially this gives us 5 ways to succeed (since there are 5 red pencils) out of 12 total things to do (since there are 5 + 3 + 4 = 12 pencils). That's 5/12 for the first pencil.

But we need to pick two more pencils! Now that we've gotten a red out, we have to recalculate the chance of pulling it. There are 4 ways to succeed (since there are 4 red left) out of 11 possible outcomes (for each pencil). That's 4/11.

For the third time, there are 3 ways to succeed out of 10 outcomes. That's 3/10.

Since we need all three things to happen, that means that we multiply the probabilities:
$P \left(\textrm{3 red}\right) = P \left(\textrm{1 red}\right) \cdot P \left(\textrm{2 red s a f t e r 1 red}\right) \cdot P \left(\textrm{3 r d red}\right)$
$= \frac{5}{12} \cdot \frac{4}{11} \cdot \frac{3}{10} = \frac{5 \cdot 4 \cdot 3}{12 \cdot 11 \cdot 10} = \frac{5}{11 \cdot 10} = \frac{1}{22} .$

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