Originally Posted by

**director** A fair coin is flipped 10 times. What is the probability that the heads turn up 8 or more times (8,9,10)?

My approach:

Probability that H turn up 8 times:

$\displaystyle P(8) = {10 \choose 8} (0.5)^{8} (0.5)^{2}$

Probability that H turn up 9 times:

$\displaystyle P(9) = {10 \choose 9}(0.5)^{10}$

Probability that H turn up 10 times:

$\displaystyle P(10) = {10 \choose 10} (0.5)^{10}$

I'm not 100% sure what to do now. Do I simply add the individual probabilities to get the total?

$\displaystyle \color{red}{\text{Yes.}}$

$\displaystyle P(8,9,10) = P(8) + P(9) + P(10)$

Thanks!