• Quiz: We have two six-sided dice. When they are tolled, it could end up with the following occurance: (A) dice 1 lands on side “3”, (B) dice 2 lands on side “1”, and (C) Two dice sum to eight.

Question

1) 𝑃𝑃(𝐴𝐴) = ? 2) 𝑃𝑃(𝐵𝐵) =? 3) 𝑃𝑃(𝐶𝐶) = ? 4) 𝑃𝑃(𝐴𝐴|𝐵𝐵) = ? 5) 𝑃𝑃(𝐶𝐶|𝐴𝐴) = ? 6) 𝑃𝑃(𝐴𝐴, 𝐵𝐵) = ? 7) 𝑃𝑃(𝐴𝐴, 𝐶𝐶) = ? 8) Is 𝑃𝑃(𝐴𝐴, 𝐶𝐶) equals 𝑃𝑃(𝐴𝐴) ∗ 𝑃𝑃(𝐶𝐶) ?

Answer 1

1) P(A) = 1/6, because one of die 1's six sides has a 3.

2) P(B) = 1/6, because die 2 has a single side with a 1.

3) P(C) = 5/36. Since each die has 6 sides, there are 6*6=36 possible outcomes for rolling both dice. 5 of which add to 8: "2:6","3:5","4:4","5:3","6:2".

4) P(A|B) = 1/6. The formula for conditional probability is P(A|B) = P(A,B)/P(B). The probability that die 1 lands on 3 and die 2 on 1 is 1/36. (1/36)/(1/6) = 1/6. We can intuit this answer also though, since A and B could be considered independent events. When two events A and B are independent, P(A) = P(A|B).

5) P(C|A) = 1/6. This asks the probability of getting a sum of 8, if die one has landed on 3. Since only die 2 is uncertain, we know die 2 must land on 5. Thus, 1/6.

6) P(A,B) = P(A)*P(B) = (1/6)*(1/6) = 1/36 since A and B are independent events.

7) P(A,C) = 1/36, since only one of the 36 outcomes has A=3 and the sum = 8: "3,5".

8) No. P(A,C) = 1/36, whereas P(A)*P(C) = (1/6)*(5/36) = 5/216. A and C are not independent events.