Probability of 3 or more occurences out of 10

Hi.

I know how to calculate the probability of rolling a 6 on a standard 6-sided die 3 times out of 3 etc. What I can't remember how to calculate is this:

If I were to roll TEN dice what is the probability that I would roll a 6 on:

a) EXACTLY 3 occasions

b) AT LEAST 3 occasions

If anybody can help me I would be grateful. Many thanks!

Re: Probability of 3 or more occurences out of 10

If I understand correctly, you are rolling 10 dice and you want to calculate the probability that:

a) exactly three 6's are showing.

b) at least three 6's are showing.

a) We need to know how many ways there are to choose 3 from 10:

$\displaystyle {10 \choose 3}=\frac{10!}{3!(10-3)!}=120$

We also need to know the probability of getting one particular set of 3 dice to show 6's:

$\displaystyle \left(\frac{1}{6}\right)^3 \left(\frac{5}{6}\right)^7$

Hence, the probability in question is:

$\displaystyle P(X)=120\left(\frac{1}{6}\right)^3 \left(\frac{5}{6}\right)^7=\frac{390625}{2519424}$

b) Here, it will be simpler to calculate the probability of getting less than three 6's, i.e., to get 0, 1, or 2 6's, then subtract this from 1 (Do you understand why?). See if you can use the same technique outlined in part a) to find this probability.

Re: Probability of 3 or more occurences out of 10

Hi.

Thanks for that. Think it makes sense!

Cheers :)

Re: Probability of 3 or more occurences out of 10

Hello, CJ1969!

Quote:

If I were to roll TEN dice, what is the probability that I would roll a 6 on:

b) AT LEAST 3 occasions

"At least three 6's" means:

. . three 6's, four 6's. five 6's, six 6's, seven 6's, eight 6's, nine 6's, or ten 6's.

We must calculate the probabilities of all eight cases and add.

It is easier to calculate the opposite probability: "less than three 6's".

. . $\displaystyle \begin{array}{cccccccc} \displaystyle P(\text{no 6's}) &=& {10\choose0}\left(\frac{1}{6}\right)^0\left( \frac{5}{6}\right)^{10} &=& \dfrac{5^{10}}{6^{10}} \\ \\ P(\text{one 6}) &=& {10\choose1}\left(\frac{1}{6}\right)^1\left( \frac{5}{6}\right)^9 &=& 10\cdot\dfrac{5^9}{6^{10}} \\ \\ P(\text{two 6's}) &=& {10\choose2}\left(\frac{1}{6}\right)^2\left( \frac{5}{6}\right)^8 &=& 45\cdot\dfrac{5^8}{6^{10}} \end{array} $

Add them and subtract from 1.

Re: Probability of 3 or more occurences out of 10

Well I tried to work everything I needed out from what MarkFL2 had advised and I have now done the same from Soroban's advice to come up with the answer via a different route and.......

......I got the same result!

This makes me happy!! :)

Many thanks to both of you.