Example — Biased coin

Example — Biased coin, exactly two heads

A biased coin comes up heads in $\tfrac{1}{3}$ of cases. It is tossed $5$ times: what is the probability of getting exactly $2$ heads?

We apply the binomial formula with $n=5$, $k=2$, $p=1/3$:

$P_5(2) = \binom{5}{2}\left(\frac{1}{3}\right)^2\left(1-\frac{1}{3}\right)^{5-2} = 10\cdot \frac{1}{9}\cdot \frac{8}{27} = \frac{80}{243} \approx 0,329.$

Careful

If we ask “the first two are heads and the rest tails” the answer is much smaller: only $\left(\tfrac{1}{3}\right)^2\left(\tfrac{2}{3}\right)^3 = 8/243$. The difference is the factor $\binom{5}{2}=10$: without a constraint on the order, one multiplies by the number of ways the two heads can be distributed among the five positions.

Links

Topics: Probability
Concepts: Binomial distribution
Methods: Binomial
Skills: Probability calculation