You win a certain game by either pushing a button once and getting a green light, or

pushing the button twice and getting either 2 green lights or exactly one red light. The

probability p > 0 of a green light on a single push is constant and three times the

probability of a red light. Find the value of p for which the probability of winning is the

same for either the one-push or two-push option. (Show your work)

3/7

Step-by-step explanation:

p is the probability of a green light. p/3 is the probability of a red light.

The probability of getting 2 green lights is:

p²

The probability of getting exactly one red light is:

(1-p/3) * p/3 + p/3 * (1-p/3)

= 2/3 p (1-p/3)

= 2/3 p - 2/9 p²

(You can also show this with binomial probability.)

Therefore, the total probability of winning with two pushes is:

p² + 2/3 p - 2/9 p² = 7/9 p² + 2/3 p

If this equals the probability of winning with one push:

p = 7/9 p² + 2/3 p

1 = 7/9 p + 2/3

1/3 = 7/9 p

p = 3/7