Q. 2.3 - A First Course in Probability

Question

Two dice are thrown. Let $E$ be the event that the sum of the dice is odd, let $F$ be the event that at least one of the dice lands on $1$, and let $G$ be the event that the sum is $5$. Describe the events $EF, E\cup F,FG, E{F}^c, and EFG$.

Step-by-Step Solution

Answer

$$\begin{gathered} EF = \left\{(1,2),(1,4),(1,6),(2,1),(4,1),(6,1)\right\} \\ E\cup F = \left\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,3),(2,5),(3,1) \\ ,(3,2),(3,4),(3,6),(4,1),(4,3),(4,5),(5,1),(5,2), \\ (5,4),(5,6),(6,1),(6,3),(6,5)\right\} \\ FG = \left\{(1,4),(4,1)\right\} \\ E{F}^c = \left\{(2,3),(2,5),(3,2),(3,4),(3,6),(4,3),(4,5),(5,2),(5,4),(5,6),(6,3),(6,5)\right\} \\ EFG = \left\{(1,4),(4,1)\right\} \end{gathered}$$

Step 1. Describe the event E F .

Let $x_1$ be the number that appears on first dice and $x_2$ be the number that appears on second dice .

$E=\left\{\left(x_1, x_2\right):x_1+ x_2=odd \right\}$

$F=\left\{\left(1, x_2\right), \left(x_1, 1\right) \right\}$

$$\begin{gathered} E=\left\{\left(1,2\right), (1,4), (1,6), (2,1), (2,3), (2,5),\left(3,2\right), (3,4), (3,6), \\ (4,1), (4,3), (4,5), \left(5,2\right), (5,4), (5,6), (6,1), (6,3), (6,5)\right\} \\ F = \left\{\left(1,1\right), (1,2), (1,3), (1,4), (1,5), (1,6),\left(2,1\right), \\ (3,1), (4,1), (5,1), (6,1)\right\} \end{gathered}$$

Therefore, $EF = \left\{(1,2),(1,4),(1,6),(2,1),(4,1),(6,1)\right\}$

Step 2. Describe the event E ∪ F .

Let $x_1$ be the number that appears on first dice and $x_2$ be the number that appears on second dice

$E=\left\{\left(x_1, x_2\right):x_1+ x_2=odd \right\}$

$F=\left\{\left(1, x_2\right), \left(x_1, 1\right) \right\}$

$$\begin{gathered} E=\left\{\left(1,2\right), (1,4), (1,6), (2,1), (2,3), (2,5),\left(3,2\right), (3,4), (3,6), \\ (4,1), (4,3), (4,5), \left(5,2\right), (5,4), (5,6), (6,1), (6,3), (6,5)\right\} \\ F = \left\{\left(1,1\right), (1,2), (1,3), (1,4), (1,5), (1,6),\left(2,1\right), \\ (3,1), (4,1), (5,1), (6,1)\right\} \end{gathered}$$

Therefore,  $$\begin{gathered} E\cup F = \left\{(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,3),(2,5),(3,1) \\ ,(3,2),(3,4),(3,6),(4,1),(4,3),(4,5),(5,1),(5,2), \\ (5,4),(5,6),(6,1),(6,3),(6,5)\right\} \end{gathered}$$

Step 3. Describe the event F G

Let $x_1$ be the number that appears on first dice and $x_2$ be the number that appears on second dice

$F=\left\{\left(1, x_2\right), \left(x_1, 1\right) \right\}$

$$\begin{gathered} F = \left\{\left(1,1\right), (1,2), (1,3), (1,4), (1,5), (1,6),\left(2,1\right), \\ (3,1), (4,1), (5,1), (6,1)\right\} \end{gathered}$$

$G=\left\{\left(x_1, x_2\right):x_1+ x_2=5 \right\}$

$G=\left\{\left(1,4\right), (2,3), (3,2), (4,1)\right\}$

Therefore, $FG = \left\{(1,4),(4,1)\right\}$

Step 4. Describe the event E F c .

Let $x_1$ be the number that appears on first dice and $x_2$ be the number that appears on second dice.

$E=\left\{\left(x_1, x_2\right):x_1+ x_2=odd \right\}$

$F=\left\{\left(1, x_2\right), \left(x_1, 1\right) \right\}$

$$\begin{gathered} E=\left\{\left(1,2\right), (1,4), (1,6), (2,1), (2,3), (2,5),\left(3,2\right), (3,4), (3,6), \\ (4,1), (4,3), (4,5), \left(5,2\right), (5,4), (5,6), (6,1), (6,3), (6,5)\right\} \\ F = \left\{\left(1,1\right), (1,2), (1,3), (1,4), (1,5), (1,6),\left(2,1\right), \\ (3,1), (4,1), (5,1), (6,1)\right\} \end{gathered}$$

$$\begin{gathered} F^c=S-F \\ =\left\{\left(2,2\right), (2,3), (2,4), (2,5), (2,6), \left(3,2\right), (3,3), (3,4), (3,5), (3,6), \\ \left(4,2\right), (4,3), (4,4), (4,5), (4,6), \left(5,2\right), (5,3), (5,4), (5,5), (5,6), \\ \left(6,2\right), (6,3), (6,4), (6,5), (6,6),\right\} \end{gathered}$$

Therefore, $E{F}^c = \left\{(2,3),(2,5),(3,2),(3,4),(3,6),(4,3),(4,5),(5,2),(5,4),(5,6),(6,3),(6,5)\right\}$

Step 5. Describe the event E F G .

Let $x_1$ be the number that appears on first dice and 

$x_2$ be the number that appears on second dice 

$E=\left\{\left(x_1, x_2\right):x_1+ x_2=odd \right\}$

$F=\left\{\left(1, x_2\right), \left(x_1, 1\right) \right\}$

$G=\left\{\left(x_1, x_2\right):x_1+ x_2=5 \right\}$

$$\begin{gathered} E=\left\{\left(1,2\right), (1,4), (1,6), (2,1), (2,3), (2,5),\left(3,2\right), (3,4), (3,6), \\ (4,1), (4,3), (4,5), \left(5,2\right), (5,4), (5,6), (6,1), (6,3), (6,5)\right\} \\ F = \left\{\left(1,1\right), (1,2), (1,3), (1,4), (1,5), (1,6),\left(2,1\right), \\ (3,1), (4,1), (5,1), (6,1)\right\} \\ G=\left\{\left(1,4\right), (2,3), (3,2), (4,1)\right\} \end{gathered}$$

Therefore, $EFG = \left\{(1,4),(4,1)\right\}$