i have a Probabilities Question :

We are to find the probability that when three dice are rolled at the same time, the largest value of the three numbers rolled is 4.

let *A* be the outcome in which the largest number is 4, let *B* be the outcome in which the largest number is 4 or less, and let *C* be the outcome in which the largest number is 3 or less.

Let *P(X)* denote the probability that the outcome of an event is *X* .Then

(1) **P(B) = ........** , ** P(C) = ..........**

(2) since **B = A U C** and the outcomes **A** and **C** are mutually exclusive it follows that **P(A) = .........**

No vote yet

2 votes

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestTo calculate P(B), think about how many ways we can choose an ordered triplet (a,b,c) such that each of a, b, c is at most 4. How many choices can we make for a? How many choices can we make for b? How many choices can we make for c? A little thought should show you that there are 4 choices each, and they are all independent, so there should be \( 4^3 = 64 \) such outcomes. Similarly, we can enumerate the outcomes corresponding to P(C); there should be \( 3^3 = 27 \). Therefore, the number of desired outcomes that have a maximum of

exactly4 is equal to 64 - 27 = 37. How many unrestricted possible outcomes are there? \( 6^3 = 216 \), so P(A) = 37/216.This question is a good example of how calculating the probability mass function for an order statistic of a discrete probability distribution is made easier by considering the cumulative distribution function of the order statistic. That is to say, \( \Pr[X_{(k)} = x] = \Pr[X_{(k)} \le x] - \Pr[X_{(k)} \le x-1] \) for a distribution whose support is a subset of the integers, and it is generally easier to compute the RHS rather than trying to enumerate the LHS directly.

Log in to reply