A fair die is rolled $(2n+1)$ times. The probability that the faces with even number show odd number of times is
- $\displaystyle \frac {3n+1}{4n+2}$
- $\displaystyle \frac {3}{4}$
- $\displaystyle \frac {2}{3}$
- $\displaystyle \frac {1}{2}$
In a fair die, odd numbers are {1, 3, 5} and even numbers are {2, 4, 6}. The probability of rolling an even number is 1/2. In (2n+1) trials, the probability of getting an odd number of successes in a binomial distribution with p=1/2 is always 1/2.
Let p be the probability of rolling an even number, which is 1/2, making q = 1/2 as well. When a die is rolled (2n+1) times, the difference between the probability of an even number appearing an odd number of times versus an even number of times is given by the binomial identity nC0(p+q)^n minus nC0(p-q)^n. Substituting the values gives (1/2 + 1/2)^(2n+1) minus (1/2 - 1/2)^(2n+1), which equals 1 minus 0. Because the total probability of all outcomes is 1, the probability of an odd number of appearances is exactly 1/2.