Question

Suppose you decided to keep flipping a coin until tails came up, at which

point you would stop. What is the probability of tails coming up on the 4th flip

of the coin? Round your answer to the nearest tenth of a percent.

A. 3.1%

B. 6.3%

C. 1.6%

D. 12.5%

Answer 1

Answer: C 1.6 APE.X approved

Answer 2

B. 6.3%

Explanation:

For each time that the coin is tosse, there are only two possible outcomes. Either it comes up tails, or it does not. The probability of coming up tails on a toss is independent of any other toss. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

In which
`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

Fair coin:

Equally as likely to come up heads or tails, so
`p = 0.5`

Probability that the first tails comes up on the 4th flip of the coin?

0 tails during the first three, which is P(X = 0) when n = 3.

Tails in the fourth, with probability 0.5. So

`p = 0.5P(X = 0)`

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`p = 0.5P(X = 0) = 0.5*(C_(3,0).(0.5)^(0).(0.5)^(3)) = 0.0625`

0.0625 * 100 = 6.25%

Rounding to the nearest tenth of a percent, the correct answer is:

B. 6.3%