Rolling 3 dice

• May 10th 2009, 03:54 AM
sanv
Rolling 3 dice
Hi,

I have the following question to answer:

"Three fair dice are rolled. What is the probability of at least one of the dice coming up a four, five or a six?"

I calculated it the following way:

P(not 4 or 5 or 6 coming up on 1 dice) = 3/6 = 1/2

for 3 dice = 1/2 * 1/2 * 1/2 = 1/8

P(4 or 5 or 6 coming up on any of 3 dice) = 1 - P(not 4 or 5 or 6 on any of 3 dice) = 1 - 1/8 = 7/8

Is this approach correct?

Can this also be calculated with the binomial distribution? Could you please give a solution to that?

Thanks.
• May 10th 2009, 04:07 AM
Plato
Quote:

Originally Posted by sanv
"Three fair dice are rolled. What is the probability of at least one of the dice coming up a four, five or a six?"
P(not 4 or 5 or 6 coming up on 1 dice) = 3/6 = 1/2
for 3 dice = 1/2 * 1/2 * 1/2 = 1/8
P(4 or 5 or 6 coming up on any of 3 dice) = 1 - P(not 4 or 5 or 6 on any of 3 dice) = 1 - 1/8 = 7/8

Can this also be calculated with the binomial distribution? Could you please give a solution to that?

Yes, that approach is correct.

For the binomial distribution: $\sum\limits_{k = 1}^3 {\binom{3}{k}\frac{1}{{2^3 }}}$