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Consider a map of a city like the one represented in Figure 1. A person at the i

ID: 3204115 • Letter: C

Question

Consider a map of a city like the one represented in Figure 1. A person at the intersection between 1^st and 1st Avenue flips a fair coin. If the outcome is Head, the person moves one block in direction North, otherwise, if the outcome is Tail, the person moves one block in direction East. Then, the person flips the coin again and walks one block in direction North if the outcome is Head or one block in direction East if the outcome is Tail, and so on... What is the probability that the person sooner or later reaches the intersection between 6^th Avenue and 4^th Street? What is the probability that the person sooner or later reaches the intersection between 6^th Avenue and 4^th Street having passed through the intersection between 3^rd Avenue and 3^rd Street? What is the probability that, given that the person has reached the intersection between 6^th Avenue and 4^th Street, s/he has passed through the intersection between 3^rd Avenue and 3^rd Street?

Explanation / Answer

A) As the person can move only North or east, the person can reach the intersection between 6th Avenue and 4th Street only in exactly 9 moves of which 3 should be in the North Direction.

The probability can be given as

9C3 (0.5)^3 (0.5)^6 = 0.1641

B) The person has to be at the intersection between 3rd Avenue and 3rd Street

As the person can move only North or east, the person can reach the intersection between 3rd Avenue and 3rd Street only in exactly 5 moves of which 2 should be in the North Direction.

Further the person has to reach the intersection between 6th Avenue and 4th Street

As the person can move only North or east, the person can reach the intersection between 6th Avenue and 4th Street only in exactly 4 moves of which 1 should be in the North Direction.

The combined probability is given by

5C2 (0.5)^2 (0.5)^3 * 4C1 (0.5)^1 (0.5)^3 = 0.0781

C) This will be same as the probability in B)

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