Binomial trial, example

Question

I have a rather standard problem. Suppose that 10 unfair coins are tossed. What is the probability of obtaining 4 successes (H) where the probability of success (H) is 6/7 and failure (T) is 1/7? I'm looking for a symbolic formula involving binomial coefficients.

Answer 1

If the coin would be fair then the answer would be:$$\binom{10}4\left(\frac12\right)^4\left(\frac12\right)^6=2^{-10}\binom{10}4$$

This corresponds with the fact that $$\sum_{k=0}^{10}\binom{10}k1^k1^{10-k}=\left(1+1\right)^{10}=2^{10}$$

In this case the answer is $$\binom{10}4\left(\frac67\right)^4\left(\frac17\right)^6=7^{-10}\binom{10}46^4$$

This corresponds with the fact that $$\sum_{k=0}^{10}\binom{10}k6^k1^{10-k}=\left(6+1\right)^{10}=7^{10}$$

Maybe this makes things clear.

Answer 2

The answer is $$\binom {10} {4} (6/7)^4 (1/7)^6$$