Question

Given that in 4 flips of a fair coin there are at least two "heads", what is the probability that there are two "tails"? There are ten equally likely outcomes: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. You randomly select one value, call it the initial value. Then, you continue to randomly select values, call them follow-up selections, until you come up with the initial value. What is the fewest number of follow-up selections that insures that the probability is better than one-half that you will observe your initial value? (Note: this problem assumes values are selected "with replacement," which simply means that after each selection, there are still the same ten equally likely outcomes.)

Answer 1

The the number of tails/heads out of 4 flips of a fair coin has Binomial distribution with
The PMF of is .

The probability of at least two "heads" is

The probability that there are two "tails"

The number of follow-up selections needed to observe the initial value has geometric distribution.

The PMF is

The probability  that you will observe your initial value in selections is

We need,

Taking logarithms on both sides (Here logarithms results in negative value. So the inequality sign reverses after division by negative number)

The minimum number of follow-up selections is .