Chapter 2 Section 2 Question 4
Question
Suppose we roll one fair six-sided die, and let Z be the number showing. Let $W = Z^3 + 4$, and let $V = \sqrt{Z}$.
  1. Compute $P(W = w)$ for every real number w
  2. Compute $P(V = v)$ for every real number v.
  3. Compute $P(ZW = x)$ for every real number x.
  4. Compute $P(VW = y)$ for every real number y.
  5. Compute $P(V + W = r )$ for every real number r.
Answer
Author: Mohammad-Ali Bandzar| Date:Oct 14 2020
(a) Compute $P(W = w)$ for every real number w
$$Z=1$$ $$W(1) = (1)^3 + 4$$ $$W(1)=5$$ $$P(W=5)=P(Z=1)=\frac{1}{6}$$ $$Z=2$$ $$W(2) = (2)^3 + 4$$ $$W(2)=12$$ $$P(W=12)=P(Z=2)=\frac{1}{6}$$ $$Z=3$$ $$W(3) = (3)^3 + 4$$ $$W(3)=31$$ $$P(W=31)=P(Z=3)=\frac{1}{6}$$ $$Z=4$$ $$W(4) = (4)^3 + 4$$ $$W(4)=68$$ $$P(W=68)=P(Z=4)=\frac{1}{6}$$ $$Z=5$$ $$W(5) = (5)^3 + 4$$ $$W(5)=129$$ $$P(W=129)=P(Z=5)=\frac{1}{6}$$ $$Z=6$$ $$W(6) = (6)^3 + 4$$ $$W(6)=220$$ $$P(W=220)=P(Z=6)=\frac{1}{6}$$ The probability of any other w is zero $P(W=w)=0$ for $w\notin \{5,12,31,68,129,220\}$
(b) Compute $P(V = v)$ for every real number v.
$$V(1) = \sqrt{1}$$ $$V(1)=1$$ $$P(V=1)=P(Z=1)=\frac{1}{6}$$ $$V(2) = \sqrt{2}$$ $$P(V=\sqrt{2})=P(Z=2)=\frac{1}{6}$$ $$V(3) = \sqrt{3}$$ $$P(V=\sqrt{3})=P(Z=3)=\frac{1}{6}$$ $$V(4) = \sqrt{4}$$ $$P(V=\sqrt{4})=P(Z=4)=\frac{1}{6}$$ $$V(5) = \sqrt{5}$$ $$P(V=\sqrt{5})=P(Z=5)=\frac{1}{6}$$ $$V(6) = \sqrt{6}$$ $$P(V=\sqrt{6})=P(Z=6)=\frac{1}{6}$$ The probability of any other v is zero $P(V=v)=0$ for $v\notin \{1,\sqrt{2},\sqrt{3},\sqrt{4},\sqrt{5},\sqrt{6}\}$
(c) Compute $P(ZW = x)$ for every real number x.
Note: values of W will be taken from question a above $$ZW(1)=Z(1)\times W(1)$$ $$ZW(1) = 1\times 5$$ $$ZW(1)=5$$ $$P(ZW=5)=P(Z=1)=\frac{1}{6}$$ $$ZW(2)=Z(2)\times W(2)$$ $$ZW(2) = 2\times 12$$ $$ZW(2)=24$$ $$P(ZW=24)=P(Z=2)=\frac{1}{6}$$ $$ZW(3)=Z(3)\times W(3)$$ $$ZW(3) = 3\times 31$$ $$ZW(3)=93$$ $$P(ZW=93)=P(Z=3)=\frac{1}{6}$$ $$ZW(4)=Z(4)\times W(4)$$ $$ZW(4) = 4\times 68$$ $$ZW(4)=272$$ $$P(ZW=24)=P(Z=4)=\frac{1}{6}$$ $$ZW(5)=Z(5)\times W(5)$$ $$ZW(5) = 5\times 129$$ $$ZW(5)=645$$ $$P(ZW=645)=P(Z=5)=\frac{1}{6}$$ $$ZW(6)=Z(6)\times W(6)$$ $$ZW(6) = 6\times 220$$ $$ZW(6)=1320$$ $$P(ZW=1320)=P(Z=6)=\frac{1}{6}$$ The probability of any other zw is zero $P(ZW=x)=0$ for $x\notin \{5,24,93,272,645,1320\}$
(d) Compute $P(VW = y)$ for every real number y.
Note: values of W and V will be taken from question a above $$VW(1) = V(1)\times W(1)$$ $$VW(1)=1\times 5$$ $$VW(1)=5$$ $$P(VW=5)=P(Z=1)=\frac{1}{6}$$ $$VW(2) = V(2)\times W(2)$$ $$VW(2)=\sqrt{2}\times 12$$ $$VW(2)=12\sqrt{2}$$ $$P(VW=12\sqrt{2})=P(Z=2)=\frac{1}{6}$$ $$VW(3) = V(3)\times W(3)$$ $$VW(3)=\sqrt{3}\times 31$$ $$VW(3)=31\sqrt{3}$$ $$P(VW=31\sqrt{3})=P(Z=3)=\frac{1}{6}$$ $$VW(4) = V(4)\times W(4)$$ $$VW(4)=\sqrt{4}\times 68$$ $$VW(4)=68\sqrt{4}$$ $$P(VW=68\sqrt{4})=P(Z=4)=\frac{1}{6}$$ $$VW(5) = V(5)\times W(5)$$ $$VW(5)=\sqrt{5}\times 129$$ $$VW(5)=129\sqrt{2}$$ $$P(VW=129\sqrt{2})=P(Z=5)=\frac{1}{6}$$ $$VW(6) = V(6)\times W(6)$$ $$VW(6)=\sqrt{6}\times 220$$ $$VW(6)=220\sqrt{6}$$ $$P(VW=220\sqrt{2})=P(Z=6)=\frac{1}{6}$$ The probability of any other vw is zero $P(VW=y)=0$ for $y\notin \{5,12\sqrt{2},31\sqrt{3},68\sqrt{4},129\sqrt{5},220\sqrt{6}\}$
(e) Compute $P(V + W = r )$ for every real number r.
Note: values of W and V will be taken from question a above $$V(1)+W(1) = V(1)+ W(1)$$ $$V(1)+W(1)=1+ 5$$ $$V(1)+W(1)=6$$ $$P(V+W=6)=P(Z=1)=\frac{1}{6}$$ $$V(2)+W(2) = V(2)+ W(2)$$ $$V(2)+W(2)=\sqrt{2}+ 12$$ $$P(V+W=\sqrt{2}+ 12)=P(Z=2)=\frac{1}{6}$$ $$V(3)+W(3) = V(3)+ W(3)$$ $$V(3)+W(3)=\sqrt{3}+ 31$$ $$P(V+W=\sqrt{3}+ 31)=P(Z=3)=\frac{1}{6}$$ $$V(4)+W(4) = V(4)+ W(4)$$ $$V(4)+W(4)=\sqrt{4}+ 68$$ $$P(V+W=\sqrt{4}+ 68)=P(Z=4)=\frac{1}{6}$$ $$V(5)+W(5) = V(5)+ W(5)$$ $$V(5)+W(5)=\sqrt{5}+ 129$$ $$P(V+W=\sqrt{5}+ 129)=P(Z=5)=\frac{1}{6}$$ $$V(6)+W(6) = V(6)+ W(6)$$ $$V(6)+W(6)=\sqrt{6}+ 220$$ $$P(V+W=\sqrt{6}+ 220)=P(Z=6)=\frac{1}{6}$$ The probability of any other v+w is zero $P(V+W=r)=0$ for $r\notin \{6,\sqrt{2}+ 12,\sqrt{3}+ 31,\sqrt{4}+ 68,\sqrt{5}+ 129,\sqrt{6}+ 220\}$