Probability & Statistics Ch.2 Section 2 Question 4

Solutions for “Probability and Statistics: The Science of Uncertainty” (Second Edition). These are solutions I have come up with; I offer no guarantee of accuracy.

Question

Suppose we roll one fair six-sided die, and let Z be the number showing. Let $W = Z^3 + 4$, and let $V = \sqrt{Z}$.

a) Compute $P(W = w)$ for every real number w b) Compute $P(V = v)$ for every real number v c) Compute $P(ZW = x)$ for every real number x d) Compute $P(VW = y)$ for every real number y e) Compute $P(V + W = r)$ for every real number r

Solution

(a) Compute $P(W = w)$ for every real number w

For each value of Z:

• $Z=1$: $W(1) = 1^3 + 4 = 5$, so $P(W=5) = \frac{1}{6}$
• $Z=2$: $W(2) = 2^3 + 4 = 12$, so $P(W=12) = \frac{1}{6}$
• $Z=3$: $W(3) = 3^3 + 4 = 31$, so $P(W=31) = \frac{1}{6}$
• $Z=4$: $W(4) = 4^3 + 4 = 68$, so $P(W=68) = \frac{1}{6}$
• $Z=5$: $W(5) = 5^3 + 4 = 129$, so $P(W=129) = \frac{1}{6}$
• $Z=6$: $W(6) = 6^3 + 4 = 220$, so $P(W=220) = \frac{1}{6}$

$P(W=w)=0$ for $w\notin \{5,12,31,68,129,220\}$

(b) Compute $P(V = v)$ for every real number v

• $V(1) = \sqrt{1} = 1$, so $P(V=1) = \frac{1}{6}$
• $V(2) = \sqrt{2}$, so $P(V=\sqrt{2}) = \frac{1}{6}$
• $V(3) = \sqrt{3}$, so $P(V=\sqrt{3}) = \frac{1}{6}$
• $V(4) = \sqrt{4} = 2$, so $P(V=2) = \frac{1}{6}$
• $V(5) = \sqrt{5}$, so $P(V=\sqrt{5}) = \frac{1}{6}$
• $V(6) = \sqrt{6}$, so $P(V=\sqrt{6}) = \frac{1}{6}$

$P(V=v)=0$ for $v\notin \{1,\sqrt{2},\sqrt{3},2,\sqrt{5},\sqrt{6}\}$

(c) Compute $P(ZW = x)$ for every real number x

• $ZW(1) = 1 \times 5 = 5$, so $P(ZW=5) = \frac{1}{6}$
• $ZW(2) = 2 \times 12 = 24$, so $P(ZW=24) = \frac{1}{6}$
• $ZW(3) = 3 \times 31 = 93$, so $P(ZW=93) = \frac{1}{6}$
• $ZW(4) = 4 \times 68 = 272$, so $P(ZW=272) = \frac{1}{6}$
• $ZW(5) = 5 \times 129 = 645$, so $P(ZW=645) = \frac{1}{6}$
• $ZW(6) = 6 \times 220 = 1320$, so $P(ZW=1320) = \frac{1}{6}$

$P(ZW=x)=0$ for $x\notin \{5,24,93,272,645,1320\}$

(d) Compute $P(VW = y)$ for every real number y

• $VW(1) = 1 \times 5 = 5$, so $P(VW=5) = \frac{1}{6}$
• $VW(2) = \sqrt{2} \times 12 = 12\sqrt{2}$, so $P(VW=12\sqrt{2}) = \frac{1}{6}$
• $VW(3) = \sqrt{3} \times 31 = 31\sqrt{3}$, so $P(VW=31\sqrt{3}) = \frac{1}{6}$
• $VW(4) = 2 \times 68 = 136$, so $P(VW=136) = \frac{1}{6}$
• $VW(5) = \sqrt{5} \times 129 = 129\sqrt{5}$, so $P(VW=129\sqrt{5}) = \frac{1}{6}$
• $VW(6) = \sqrt{6} \times 220 = 220\sqrt{6}$, so $P(VW=220\sqrt{6}) = \frac{1}{6}$

$P(VW=y)=0$ for $y\notin \{5,12\sqrt{2},31\sqrt{3},136,129\sqrt{5},220\sqrt{6}\}$

(e) Compute $P(V + W = r)$ for every real number r

• $V(1)+W(1) = 1 + 5 = 6$, so $P(V+W=6) = \frac{1}{6}$
• $V(2)+W(2) = \sqrt{2} + 12$, so $P(V+W=\sqrt{2}+12) = \frac{1}{6}$
• $V(3)+W(3) = \sqrt{3} + 31$, so $P(V+W=\sqrt{3}+31) = \frac{1}{6}$
• $V(4)+W(4) = 2 + 68 = 70$, so $P(V+W=70) = \frac{1}{6}$
• $V(5)+W(5) = \sqrt{5} + 129$, so $P(V+W=\sqrt{5}+129) = \frac{1}{6}$
• $V(6)+W(6) = \sqrt{6} + 220$, so $P(V+W=\sqrt{6}+220) = \frac{1}{6}$

$P(V+W=r)=0$ for $r\notin \{6,\sqrt{2}+12,\sqrt{3}+31,70,\sqrt{5}+129,\sqrt{6}+220\}$