An Introduction to Mathematical Statistics and Its Applications (6th Edition)

Published by Pearson
ISBN 10: 0-13411-421-3
ISBN 13: 978-0-13411-421-7

Chapter 3 Random Variables - 3.3 Discrete Random Variables - Questions - Page 126: 3

Answer

$\color{blue}{p_X(k) = (k^3-(k-1)^3)/216,\ k=1,2,3,4,5,6.}$

Work Step by Step

The sample space for the faces showing in the toss of three fair dice is $\{(i,j,k)\ \vert\ i,j,k \in \{1,2,3,4,5,6\}\}$ as each die can come up either a $1,2,3,4,5$, or $6.$ The sample space has $6^3 = 216$ elements. If the largest of the three faces that appear in the throw of three dice is $k$, then $k \in \{1,2,3,4,5,6\}$. The total number of ways in which the largest of the three faces that appear in the throw of three dice is $k$, is equal to the number of ways in which the face that appears on each of the three dice is at most $k$ less the number of ways in which the face that appears on each of the three dice is at most $k-1$. The former event can occur in $k^3$ ways, as each of three dice can come up in $k$ ways (one of $1,2,\ldots,k$), while the latter event can occur in $(k-1)^3$ ways, as each of the three dice can up in $k-1$ ways (one of $1,2,\ldots, k-1$). Therefore, the total number of ways in which the largest of the three faces that appear in the throw of three dice is $k$, is $k^3-(k-1)^3,\ k=1,2,3,4,5,6$. Thus, if we let $X$ be the random variable for the largest of the three faces that appear in the throw of three dice, $P(X=k) = (k^3-(k-1)^3)/216$ since the total number of successful events is $k^3-(k-1)^3$, while the total sample space has $216$ elements. Consequently, the pdf of $X$ is $\begin{align*} p_X(k) &= P(X=k),\ k=1,2,3,4,5,6 \\ \color{blue}{p_X(k)} &\color{blue}{= (k^3-(k-1)^3)/216,\ k=1,2,3,4,5,6}. \end{align*}$
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