## GMAT **Probability per Distribution and Equal Likelihoods**

*A bag contains red and blue balls. Exactly half of the balls are red, and the other half are blue. If four balls are removed from the bag, then replaced, what is the Probability that exactly three blue balls** will be among those selected?*

First, let’s figure out the *number of distributions.*

*P(B)*P(B)*P(B)*P(R)*

Well, it’s four total and three repeated, so by Mississippi we see four total distributions:

4!/3!=4

Next, let’s look at *Prob. per distribution, *which, given the fact that **the balls are replaced, **is actually rather straightforward. Remember, it’s a 50% chance of getting either:

*P(B)***P(B)***P(B)***P(R)*

(*1/**2)(**1/**2)(**1/**2)(**1/**2)**=**1**16*

Now all we need to do is multiply the two values we’ve found so far:

number of distributions * prob./distribution=4*(1/16)=3/8

See? Not so bad after all.

Interested in learning more about **GMAT Probability?** Check out the main resource page at the link. There’s more prob questions and answers than you can shake a stick at!

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