# Probability Trouble

• Sep 5th 2012, 12:15 AM
Mwayne95
Probability Trouble
So here's the problem:

A five-sided die has the numbers 1-5 on each side. What is the probability, after 6 rolls, of rolling a 1 or a 2 four times only?

I could do the problem if:

It only dealt with one side of the die instead of two
It didn't specifically ask for four times, but the probability of a 1 or a 2 occurring at all.

Thank you so much!
• Sep 5th 2012, 01:25 AM
kalyanram
Re: Probability Trouble
Discard this argument I messed up the calculation of probabilities follow the argument by Plato

Every roll has 5 possible outcomes therefore total number of outcomes = $\displaystyle 5^6$
No. of ways of having 4 ones is = $\displaystyle \binom{6}{4}.4^2$ which is also the case with having 4 two only as these two events are mutually exclusive we have the probability = $\displaystyle \frac{2.\binom{6}{4}.4^2}{5^6}$
• Sep 5th 2012, 02:00 AM
Plato
Re: Probability Trouble
Quote:

Originally Posted by Mwayne95
So here's the problem:
A five-sided die has the numbers 1-5 on each side. What is the probability, after 6 rolls, of rolling a 1 or a 2 four times only?

This is a simple binomial probability question. There are six trials wanting four successes.
$\displaystyle \binom{6}{4}{\left( {\frac{2}{5}} \right)^4}{\left( {\frac{3}{5}} \right)^2}$.