I'm flipping an unfair coin 5 times. The probability of getting heads is $\frac{2}{3}$ and the probability of getting tails is $\frac{1}{3}$. What's the probability that at least 3 of the coins end up being heads?
My working:
If I flip 5 times, there are $2^5 = 32$ possible outcomes.
I know that there are $16$ outcomes where there are at least 3 heads (1 with 5H, 5 with 4H, and 10 with 3H).
The probability of getting 0H is $\left(\frac{1}{3}\right)^5$. Can I extend this logic for the probability of getting 1H and 2H and simply sum those numbers and subtract from 1?
Best Answer
A coin flip follows a binomial distribution. \begin{align*} P(H\geq3)&=P(H=3)+P(H=4)+P(H=5) \\ &=\binom{5}{3}\left(\frac{2}{3}\right)^3\left(\frac{1}{3}\right)^2+\binom{5}{4}\left(\frac{2}{3}\right)^4\left(\frac{1}{3}\right)^1+\binom{5}{5}\left(\frac{2}{3}\right)^5\left(\frac{1}{3}\right)^0\\ &=\frac{64}{81} \end{align*}