# Simple Laws of Basic Probability using Tables

## Question 1

Students were required to choose either Music or Art and Biology or Chemistry.

$$\begin{array}{|c|c|c|} \hline & \ \ \ \text{Art} \ \ \ & \text{Music} \\ \hline \text{Biology} & 7 & 4 \\ \hline \text{Chemistry} & 6 & 11 \\ \hline \end{array}$$

A student is chosen at random. What is the probability that he/she:

(a)     chose Art and Chemistry?

\begin{align} \displaystyle \frac{6}{28} = \frac{3}{14} \end{align}

(b)     did not choose Music?

\begin{align} \displaystyle \text{There are } 15 \text{ students for Music, so } 13 \text{ students not for Music, so } \frac{13}{28} \end{align}

(c)     chose Biology?

\begin{align} \displaystyle \text{There are } 11 \text{ students chose Biology, so } \frac{11}{28} \end{align}

(d)     did not choose Chemistry or Art?

\begin{align} \displaystyle \text{Did not choose Chemistry or Art} &= \text{Biology} \\ \therefore \frac{4}{28} &= \frac{1}{7} \end{align}

What is the probability that a student:

(e)     who chose Biology, also, does Art?

\begin{align} \displaystyle \Pr\left( \frac{\text{Biology & Art}}{\text{Biology}} \right) = \frac{7}{11} \end{align}

(f)     who does not do Art does study Chemistry?

\begin{align} \displaystyle \Pr\left( \frac{\text{Chemistry & Music}}{\text{Music}} \right) = \frac{11}{15} \end{align}

## Question 2

A player rolls two dice with the results as shown in the table. On the second roll, the player must roll the same sum as they rolled in the first roll to continue.

$$\begin{array}{|c|c|} \hline \bbox[yellow,3px]{\text{sum of } 7 \text{ or } 11} & \text{win} \\ \hline \bbox[yellow,3px]{\text{sum of } 2, 3 \text{ or } 12} & \text{lose} \\ \hline \bbox[yellow,3px]{\text{any other sum}} & \text{roll again} \\ \hline \end{array}$$

What is the probability of:

(a)     winning on the first roll?

$$\require{AMSsymbols} \begin{array}{|r|r|r|r|r|r|r|} \hline & \ \bbox[yellow,3px]1 & \ \bbox[yellow,3px]2 & \ \bbox[yellow,3px]3 & \ \bbox[yellow,3px]4 & \ \bbox[yellow,3px]5 & \ \bbox[yellow,3px]6 \\ \hline \ \bbox[yellow,3px]1 & 2 & 3 & 4 & 5 & 6 & \bbox[pink,3px]7 \\ \hline \ \bbox[yellow,3px]2 & 3 & 4 & 5 & 6 & \bbox[pink,3px]7 & 8 \\ \hline \ \bbox[yellow,3px]3 & 4 & 5 & 6 & \bbox[pink,3px]7 & 8 & 9 \\ \hline \ \bbox[yellow,3px]4 & 5 & 6 & \bbox[pink,3px]7 & 8 & 9 & 10 \\ \hline \ \bbox[yellow,3px]5 & 6 & \bbox[pink,3px]7 & 8 & 9 & 10 & \bbox[pink,3px]{11} \\ \hline \ \bbox[yellow,3px]6 & \bbox[pink,3px]7 & 8 & 9 & 10 & \bbox[pink,3px]{11} & 12 \\ \hline \end{array}$$

\begin{align} \displaystyle \Pr (\text{sum of } 7 \text{ or } 11) = \frac{8}{36} = \frac{2}{9} \end{align}

(b)     losing on the first roll?

$$\require{AMSsymbols} \begin{array}{|r|r|r|r|r|r|r|} \hline & \ \bbox[yellow,3px]1 & \ \bbox[yellow,3px]2 & \ \bbox[yellow,3px]3 & \ \bbox[yellow,3px]4 & \ \bbox[yellow,3px]5 & \ \bbox[yellow,3px]6 \\ \hline \ \bbox[yellow,3px]1 & \bbox[pink,3px]2 & \bbox[pink,3px]3 & 4 & 5 & 6 & 7 \\ \hline \ \bbox[yellow,3px]2 & \bbox[pink,3px]3 & 4 & 5 & 6 & 7 & 8 \\ \hline \ \bbox[yellow,3px]3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \hline \ \bbox[yellow,3px]4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline \ \bbox[yellow,3px]5 & 6 & 7 & 8 & 9 & 10 & 11 \\ \hline \ \bbox[yellow,3px]6 & 7 & 8 & 9 & 10 & 11 & \bbox[pink,3px]{12} \\ \hline \end{array}$$

\begin{align} \displaystyle \Pr (\text{sum of } 2,3 \text{ or } 12) = \frac{4}{36} = \frac{1}{9} \end{align}

(c)     neither winning nor losing on the first throw?

\begin{align} \displaystyle \Pr (\text{neither}) &= 1-\Pr(\text{win})-\Pr(\text{lose}) \\ &= 1-\frac{2}{9}-\frac{1}{9} \\ &= \frac{6}{9} \\ &= \frac{2}{3} \end{align}

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