A lopsided coin has a probability of 1/3 for coming up heads and 2/3 for coming up tails. On average, how many flips of this coin are needed to have both heads and tails appear at least once? Give your answer as a reduced fraction.

Answer with explanation:For a Lopsided coin ,probability of getting Head is equal to [tex]\frac{1}{3}[/tex] For a Lopsided coin ,probability of getting Tail  is equal to [tex]\frac{1}{3}[/tex].→Probability of getting Tail > Probability of getting Head→Coin is heavier from tail side and lighter from Head side.→→We have to Calculate number of flips of coin that is needed to have both heads and tails appear at least once.[tex]\rightarrow P(\text{Head})=\frac{1}{3}\\\\ \frac{1}{3} \times x=1\\\\x=3\\\\\rightarrow P(\text{Tail})=\frac{2}{3}\\\\ \frac{2}{3} \times y=1\\\\y=\frac{3}{2}[/tex]→We need to find common multiple of 3 and [tex]\frac{3}{2}[/tex].Least common multiple of 3 and [tex]\frac{3}{2}[/tex] is 6.→So,on Average number of flips of this coin are needed to have both heads and tails appear at least once=6 tosses