Probability of a Single Event

When rolling an unbiased die, there are \(6\) possible outcomes:

1, 2 , 3, 4, 5 or 6 so \(n\begin{pmatrix}U\end{pmatrix} = 6\).

We can define an event \(A\) as: \[A: \ \text{rolling a 5}\] Only one face of the die shows the number \(5\), so the number of ways event \(A\) can occur is \(1\), \(n\begin{pmatrix}A\end{pmatrix} = 1\). The probability of rolling a \(5\) is therefore: \[p\begin{pmatrix}A\end{pmatrix} = \frac{n\begin{pmatrix}A\end{pmatrix}}{n\begin{pmatrix}U\end{pmatrix}}\] That's: \[p\begin{pmatrix}A\end{pmatrix} = \frac{1}{6} \approx 17\%\] The probability of rolling a \(5\) is therefore \(p\begin{pmatrix}A\end{pmatrix} = \frac{1}{6}\), which is (approximately) a \(17\%\) chance.

Since the number is picked amongst \(20\) whole numbers, the total number of possible outcomes is \(20\). So we can write \(n\begin{pmatrix}U\end{pmatrix} = 20\).

The prime numbers between \(1\) and \(20\), included, are:

2, 3, 5, 7, 11, 13, 17 and 19

Since there are \(8\) prime numbers between \(1\) and \(20\), the number of wauys event \(A\) can occur is: \(n\begin{pmatrix}A\end{pmatrix} = 8\) and we can calculate the probability: \[\begin{aligned} p\begin{pmatrix}A\end{pmatrix} & = \frac{n\begin{pmatrix}A\end{pmatrix}}{\begin{pmatrix}U\end{pmatrix}} \\ & = \frac{8}{20} \\ & = \frac{2}{5} \\ p\begin{pmatrix}A\end{pmatrix} & = 0.4 \end{aligned}\] The probability that the number be prime is \(p\begin{pmatrix}A\end{pmatrix} = 0.4\).

This shows us clearly that there are \(4\) outcomes in this experment. Indeed the spinner will stop on either of the 4 quarters: White, Blue, Orange, Orange So: \[n\begin{pmatrix}U\end{pmatrix} = 4\] We now answer the questions.

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1. We can define the event \(B\) as: \[B: \ \text{the spinner stops on Blue}\] Since there is "only" one Blue region the number of outcomes corresponding to Blue is \(n\begin{pmatrix}B\end{pmatrix}=1\). We can therefore state: \[\begin{aligned} p\begin{pmatrix}B\end{pmatrix} & = \frac{n\begin{pmatrix} B\end{pmatrix}}{n\begin{pmatrix}U\end{pmatrix}}\\ & = \frac{1}{4} \\ p\begin{pmatrix}B\end{pmatrix} & = 0.25 \end{aligned}\] The probability of the spinner sopping on blue is therefore \(p\begin{pmatrix}B\end{pmatrix} = 0.25\).

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2. We can define the event \(O\) as: \[O: \ \text{the spinner stops on Orange}\] Sincer there are 2 Orange regions (2 quarters) the number of outcomes corresponding to the event Orange, \(O\), is \(n\begin{pmatrix}O\end{pmatrix}=2\). So: \[\begin{aligned} p\begin{pmatrix}O\end{pmatrix} & = \frac{n\begin{pmatrix} O\end{pmatrix}}{n\begin{pmatrix}U\end{pmatrix}}\\ & = \frac{2}{4} \\ & = \frac{1}{2} \\ p\begin{pmatrix}O\end{pmatrix} & = 0.5 \end{aligned}\] The probability of the spinner sopping on orange is therefore \(p\begin{pmatrix}O\end{pmatrix} = 0.5\).