What is the probability of rolling at least 3 2s OR at least 3 6s in 6 rolls?

Question

I now understand how I can calculate exactly 3 2s and 3 6s in 6 rolls, but how do I approach the 'At least' and 'or' conditions of this question?

Answer 1

There are $5^{6-k}\binom6k$ ways to have exactly $k$ $2$s, and the same for $6$s, so that gives a total of

$$ 2\sum_{k=3}^65^{6-k}\binom6k=2(125\cdot20+25\cdot15+5\cdot6+1)=5812\;. $$

But now we've double-counted the $\binom63=20$ ways of having three $2$s and three $6$s, so the actual total is $5812-20=5792$.

This could be done ad hoc because there was only one way to fulfil both conditions; to more systematically solve more complicated problems of this sort, you can use inclusion-exclusion.

Answer 2

You generally need to add up all propabilities which ''at least'' or ''at most'' refer to.

Probability of getting exactly $k$ successes in $n$ trials is given by the formula:

$$ \frac{n!}{k!\cdot (n-k)!} p^k (1-p)^{n-k} \text{,} $$ where $p$ stands for probability of success in each trial.

If you want to calculate the probability of getting at least $3$ twos in 6 rolls you need sum the above formula from $k=3$ to $k=6$, so effectively you'll get:

$$ \sum_{k=3}^6 \frac{6!}{k!\cdot (6-k)!} \left(\frac{1}{6}\right)^k \left(\frac{5}{6}\right)^{6-k} \text{.} $$