Question

You roll two 6-sided dice numbered 1 through 6. Let A be the event that the first die shows the number 3, let B be the event that the second die shows a 5, and let E be the event that the sum of the two numbers showing is even.

Compute P(A)and P(B)and then compute P(AlB). What does this tell you about events A and B?Hint: Remember that the sample space has 36 outcomes!

Compute P(ElA)and compute P(E). What does this tell you about the events A and E?

Answer 1

A=  the event that the first die shows the number 3.

Since possible outcomes in a roll of one die are 1, 2, 3, 4, 5, 6. So P(A)=1/6.

B= the event that the second die shows a 5. Then P(B)=1/6.

Using Bayes theorem for probability we get,

since the outcome of one roll does not depend on the outcome of the other roll.

Here, the event that first die shows 3 and second die shows 5.

The events A, B are independent of each other.

P(A|B)=1/6=P(A)

E= the event that the sum of the two numbers showing is even.

Probability that sum of 2 numbers is even given that the first number is 3

Then the second number can be one of 1, 3, 5.

So P(E|A)=3/6=1/2.

Let A1=first number is 1, A2=first number is 2, A3=first number is 3, A4=first number is 4, A5=first number is 5

A6=first number is 6

Then P(E|A1)=P(E|A3)=P(E|A5)=3/6=1/2

P(E|A2)=P(E|A4)=P(E|A6)=3/6=1/2

P(E|A)=1/2=P(E) So E, A are independent.