I needed help on my probability Homework Suppose that five dice are rolled. (a)
ID: 2925418 • Letter: I
Question
I needed help on my probability Homework
Suppose that five dice are rolled.
(a) What is the probability that the sum of the dice is even?
One approach is to consider the last die. If the sum of the rst four dice is even, then what is
the probability that the total sum is even? What if the sum of the rst four dice is odd? Now
use the Law of Total Probability: Let B be the event that the total sum is odd and C be the
event that the sum of the rst four is even { calculate P(B) by conditioning on C and C^c.
(b) What is the probability that all five dice are different or their sum is even?
Explanation / Answer
(a) The sum being even or odd solely depends on the number of odd numbers that show up.
The sum is odd if odd number of odd numbers show up.
Since there are equal number of odd and even numbers on a dice,
Probability of getting an odd number = Probability of getting even number.
=> Probability of getting odd number of odd numbers (even number of even numbers) = Probability of getting even number of odd numbers (odd number of even numbers) = 1/2 = 0.5.
(b) There are six numbers and five different numbers can be chosen in 6P5 = 6! = 720 ways.
Probability = 720 / 65 = 0.0926.
Once again, since all possibiliites are considered, number of even number of different numbers = number of odd number of different numbers.
=> Probability that all five are different and their sum is not even = 0.0926 / 2 = 0.0463.
=> Probability that all five are different or their sum is even = 0.5 + 0.0463 = 0.5463.