\(\displaystyle \text{Tossing a head on a bent coin has a given probability of }\frac{1}{3}\)

\(\displaystyle \text{so the prob. that of tossing four heads would be: }\:1H^4T^0 \:= \:1\left(\frac{1}{3}\right)^4 \:= \:\frac{1}{81}\)

\(\displaystyle \text{ But what about this one:}\)

\(\displaystyle \text{In four tosses, what is the probability that }at\:least\:two\text{ heads will be tossed?}\)

\(\displaystyle \text{Do you have to consider: }\:1H^4T^0\,\text{ and }\,4H^3T^1\,\text{ and }\,6H^2T^2\) . . Yes!