## Monday, May 16, 2016

### Monopoly PR7: Two tries from GO.

Monopoly PR7:  Two tries from GO.

Concepts:    Probability, the union of two events, and the binomial distribution

Question:   A person rolls the two dice from Go.   After the roll of the two dice the player moves the dice back to Go and goes again.   What is the probability that at least one of the two rolls lands the person on chance, the square that is seven squares away from Go?

Hint:   The answer is NOT the sum of the probabilities from the first try and the second try.   Think of the probability of getting two heads on two coin tosses.   The answer is NOT the sum of the probabilities for the two coin tosses.

Analysis:   The probabilities from one roll of the two dice from Go were calculated in a previous post.

The probability of landing on chance is 1/6.   (Remember there are six way to get to seven with two dice (1,6), (6,1) (2,5), (5,2), (4,3), (3,4).

Define event A as the event that the person gets to Chance on the first throw.

Define event B as the event that the person gets to Chance on the second throw.

We are looking for the probability of getting to Chance on either the first or second throw, which is the union of the probability of event A and event B.

The probability of the union of events A and B is the P(A) plus P(B) minus the probability of the intersection of A and B.  Since the two throws are independent.  The probability of the intersection of A and B is the product of the two probabilities.   Our answer is:

1/6 + 1/6 – 1/36

or 11/36, which is 0.305555556.

There is another way to do this problem.   Note that the two coin tosses from Go can be described by the binomial distribution where the event of a success involves landing on Chance and the event of a failure involves not landing on Chance.   The binomial distribution for N, the number of trials equal to 2 and p the success probability equal to 1/6 are presented below.   T

 Binomial Distribution for n=2 and P=1/6 # of successes or k Probability # of Successes is k 0 0.694444444 1 0.277777778 2 0.027777778 1 or 2 0.305555556

The probability of landing on Chance on either the first or second trials is equal to the probability of landing on Chance at least once, which in this problem is once or twice.