Question 252145: A pair of dice consisting of a six-sided die and a four-sided die is rolled and the sum is determined. Let A be the event that a sum of 5 is rolled and let B be the event that a sum of 5 or a sum of 9 is rolled. Find (a) P(A), (b) P(B), and (c) P(A n B).
Found 2 solutions by edjones, drk:
Answer by edjones(8007)   (Show Source):
You can put this solution on YOUR website! a) P(5)=4/24=1/6 (1,4), (2,3), (3,2), (4,1)
b) P(9)=2/24=1/12 (5,4), (6,3) + P(5) = 4/24
2/24 + 4/24 = 6/24 = 1/4
c) P(A n B) = 0 null set
.
Ed
Answer by drk(1908)  (Show Source):
You can put this solution on YOUR website! When you roll two (2) dice, the total number of possibilities is the product of the number of sides of the first die, M, and the second die, N, or MN. In this case our total is 24.
(a) P(A) = the probability of getting the sum of 5. We can list these ways very easily in first die - second die form as: 1-4, 2-3, 3-2, 4-1. Yes you are allowed to reverse the numbers due to the different die. We see P(A) = 4/24 = 1/6.
(b) P(B) = the probability of getting the sum of 5 or 9. We can list these also very easily. We already know the sum of 5 is in 4 ways. THe sum of 9 is created in only 2 ways: 6-3, 5-4. Now, the word "OR" means add, so, P(B) = 4/24 + 2/24 = 6/24 = 1/4.
(c) P(A n B) = P(A) + P(B) - P(A u B). This is a good formula to remember. You can get it using Venn diagrams. We actually, have everything we need. It is easier not to reduce in this question until the end.
P(A n B) = 4/24 + 6/24 - (4/24 + 6/24) = 0.
It will be zero, because you can't get both the sum of a 5 AND the sum of a 9.
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