We have two six-sided dice. When they are rolled, it could end up with the following occurance: (A) dice 1 lands on side “4”, (B) dice 2 lands on side “2”, and (C) Two dice sum to seven.

Question

1- P(A)=?

2- P(B)=?

3- P(C)=?

4- P(A|B)=?

5- P(C|A)=?

6- P(A,B)=?

7- P(A,C)=?

8- Is P(A,C) equals P(A)*P(C)?

Answer 1

Let (x,y) be the outcome of two dice

x=outcome of the first die

y=outcome of the second die

(1,1),(1,2),(1,3),(1,4),(1,5),(1,6)

(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)

(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)

(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)

(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)

(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)

There are 36 total possible outcomes.

(1) Look at the fourth row of the outcomes. There are 6 possible outcomes where the first die lands on 4

P(A) = 6/36 = 1/6

(2) There are 6 possible outcomes where the second die lands on 2. (See the second column)

P(B) =6/36 =1/6

(3) The outcomes in the longest diagonal going up are the ones that add up to 7. There are also 6 of them.

P(C) = 6/36 =1/6

(4) Formula for conditional probability: P(A | B)= P(A∩B)/P(B)

A∩B = The first die lands on side 4 and second die lands on side 2

n(A∩B) = 1

∴P(A∩B) = 1/36

P(A | B) = (1/36)/(6/36) =1/6

(5) P(C | A) = P(C∩A)/P(A)

n(C∩A) = 1, (If the first die is 4 and the second die is 3).

P(C∩A) = 1/36

P(C | A) = (1/36)/(6/36) = 1/6

(6) see number (4)

(7) see number (5)

(8) P(A∩C) = 1/36

Let's check P(C∩A) = P(A)•P(C) =(1/6)(1/6) = 1/36. Yes they are equal because A and B are independent events even if they are not mutually exclusive.