Question

Johnquavious the frog is standing on lily pad 0 in a long line of infinite lily pads. Every second, he jumps forward m lily pads with a probability of 1/2^m. What is the probability he will eventually land on lily pad n (round to nearest 1e-6)?

a. 1/2^n
b. 1/2^(n-1)
c. 1/2^(n+1)
d. 1/2^(2n)

Answer 1

The probability that Johnquavious the frog will eventually land on lily pad n is 1/2^(n-1). Option B is answer.

Step-by-step explanation:

Let P(n) be the probability that Johnquavious the frog will eventually land on lily pad n.

To calculate P(n), we can consider the following cases:

Case 1: Johnquavious jumps forward m lily pads to lily pad n in one step.

The probability of this case is 1/2^m.

Case 2: Johnquavious jumps forward m lily pads to a lily pad k, where 0 < k < n, and then jumps forward m lily pads again to lily pad n.

The probability of this case is 1/2^m * P(k).

Since there are an infinite number of lily pads, we can sum over all possible values of k to get the probability of this case:

P(k) * 1/2^m = (1 - P(0)) * 1/2^m

Substituting this into the expression for P(n), we get:

P(n) = 1/2^m + (1 - P(0)) * 1/2^m

Since Johnquavious starts on lily pad 0, P(0) = 0.

Substituting P(0) = 0 into the equation, we get:

P(n) = 1/2^m + (1 - 0) * 1/2^m = 2/2^m = 1/2^(m-1)

Therefore, the probability that Johnquavious will eventually land on lily pad n is 1/2^(n-1).

In this case, n = 10 and m = 3, so the probability that Johnquavious will eventually land on lily pad 10 is:

P(10) = 1/2^(10 - 1) = 1/2^9 = 0.00438

Option B is answer.