Question

Consider the experiment where a pair of fair dice is thrown. Let X denote the random variable whose value is determined by taking the maximum of the spots showing on either of the two dice thrown. For example, if a 3 and a 5 were thrown, then X would take the value of Maximum(3,5) = 5. The range of values that X can assume are the positive integers {1,2,3,4,5,6}. Please give the corresponding probabilities for the values of X given below. Pr(X = 1) = Pr(X = 2) = Pr(X = 3) = Pr(X = 4) = Pr(X = 5) = Pr(X = 6) = Further, find the probability that X is divisible by 4. Probability that X is divisible by 4 equals ___

Answer 1

Answer:

Pr(X=1) = 1/36

Pr(X=2) = 3/36

Pr(X=3) = 5/36

Pr(X=4) = 7/36

Pr(X=5) = 9/36

Pr(X=6) = 11/36

Pr(X is divisible by 4) = 7/36

Step-by-step explanation:

Two fair die are thrown and X = maximum of the two die.

Pr(X=1) means 1 is the maximum number of the two numbers showing on the dice. This can only happen when both dice are showing 1. So,

Pr(X=1) = (1,1)

          = (1/6)*(1/6)

Pr(X=1) = 1/36

Pr(X=2) = (1,2) + (2,1) + (2,2)

            = (1/6)*(1/6) + (1/6)*(1/6) + (1/6)*(1/6)

Pr(X=2) = 3/36

Pr(X=3) = (1,3) + (2,3) + (3,1) + (3,2) + (3,3)

            = (1/6)*(1/6) + (1/6)*(1/6) + (1/6)*(1/6) + (1/6)*(1/6) + (1/6)*(1/6)

Pr(X=3) = 5/36

Pr(X=4) = (1,4) + (2,4) + (3,4) + (4,1) + (4,2) + (4,3) + (4,4)

            = (1/6)*(1/6)+(1/6)*(1/6)+(1/6)*(1/6)+(1/6)*(1/6)+(1/6)*(1/6)+(1/6)*(1/6)+(1/6)*(1/6)

Pr(X=4) = 7/36

Pr(X=5) = (1,5) + (2,5) + (3,5) + (4,5) + (5,1) + (5,2) + (5,3) + (5,4) + (5,5)

            = 9 *(1/6)*(1/6)

Pr(X=5) = 9/36

Pr(X=6) = (1,6) + (2,6) + (3,6) + (4,6) + (5,6) + (6,1) + (6,2) + (6,3) + (6,4) + (6,5) + (6,6)

            = 11 * (1/6)*(1/6)

Pr(X=6) = 11/36

X is divisible by 4 only when X=4. Other values of X are not divisible by 4. So,

Pr(X is divisible by 4) = Pr(X=4)

Pr(X is divisible by 4) = 7/36