Q.4.2 - A First Course in Probability

Question

Two fair dice are rolled. Let $X$ equal the product of the $2$ dice. Compute $P\{X=i\}\text{ for }i=1,\dots ,36$.

Step-by-Step Solution

Answer

The probability of  $P\{X=i\}$ for $i=1,\dots ,36$ will be $1$.

Step 1: Given information

Let $X$ equal the product of the $2$ dice.

Step 2: Solution

Let,

$X=$event that product of the two dice.

When two dices are rolled thye sample space will be,

$S=\{1,2,\dots 6\}\times \{1,2,\dots 6\}$

$=\left\{\begin{array}{l}1,2,\dots 6,8,9,10,12,15\\ 16,18,20,24,25,30,36\end{array}\right\}$

$\begin{array}{|lll|}\hline \text{ Outcome }& \text{ Variable value }& \text{ Probability }\\ (1,1)& X=1& P(X=1)=\frac{1}{36}\\ (1,2)(2,1)& X=2& P(X=2)=\frac{2}{36}\\ (1,3)(3,1)& X=3& P(X=3)=\frac{2}{36}\\ (4,1)(1,4)(2,2)& \mathrm{X}=4& \mathrm{P}(\mathrm{X}=4)=\frac{3}{36}\\ (1,5)(5,1)& \mathrm{X}=5& \mathrm{P}(\mathrm{X}=5)=\frac{2}{36}\\ (6,1)(1,6)(2,3),(3,2)& \mathrm{X}=6& \mathrm{P}(\mathrm{X}=6)=\frac{4}{36}\\ (2,4),(4,2)& \mathrm{X}=8& \mathrm{P}(\mathrm{X}=8)=\frac{2}{36}\\ (3,3)& \mathrm{X}=9& \mathrm{P}(\mathrm{X}=9)=\frac{1}{36}\\ (2,5)(5,2)& \mathrm{X}=10& \mathrm{P}(\mathrm{X}=10)=\frac{2}{36}\\ (2,6)(6,2)(3,4)(4,3)& \mathrm{X}=12& \mathrm{P}(\mathrm{X}=12)=\frac{4}{36}\\ (3,5)(5,3)& \mathrm{X}=15& \mathrm{P}(\mathrm{X}=15)=\frac{2}{36}\\ (4,4)& \mathrm{X}=16& \mathrm{P}(\mathrm{X}=16)=\frac{1}{36}\\ (3,6)(6,3)& \mathrm{X}=18& \mathrm{P}(\mathrm{X}=18)=\frac{2}{36}\\ (4,5)(5,4)& \mathrm{X}=20& \mathrm{P}(\mathrm{X}=20)=\frac{2}{36}\\ (4,6)(6,4)& \mathrm{X}=24& \mathrm{P}(\mathrm{X}=24)=\frac{2}{36}\\ (5,5)& \mathrm{X}=25& \mathrm{P}(\mathrm{X}=25)=\frac{1}{36}\\ (6,5)(5,6)& \mathrm{X}=30& \mathrm{P}(\mathrm{X}=30)=\frac{2}{36}\\ (6,6)& \mathrm{X}=36& \mathrm{P}(\mathrm{X}=36)=1\\ \hline\end{array}$

Total Probability $=1$

Step 3: Final answer

The probability of  $P\{X=i\}$ for $i=1,\dots ,36$ will be $1$