Probability of Single Events

This topic builds on your previous understanding of probability to include more complicated examples and guides you through the working-out process.

Example 1: Buttons

Emma has 6 pink, 16 green and 8 blue buttons in a box. What is the probability that she will randomly select a blue button? Write your answer as a percentage.

Solution:

First, calculate the total number of buttons:

  • 6 + 16 + 8 = 30 buttons in total.

Next, represent the number of blue buttons as a fraction of the total:

  • \frac{8}{30}

Finally, since the question asks for the answer in percentage form, convert the fraction to a percentage:

  • 8 \div 30 = 0.267
  • 0.267 \times 100 = 26.7\%

Example 2: Buttons (continued)

Emma removes 5 pink buttons from the box. What is the probability that she will now select a blue button? Write your answer as a decimal.

Solution:

Calculate the new total number of buttons:

  • 30 - 5 = 25

We must use this number as the new total.

Use this new total to find the fraction representing blue buttons:

  • = \frac{8}{25}

Finally, convert this fraction to a decimal:

  • 8 \div 25 = 0.32

Example 3: Numbers on a spinner

Jerry has a fair hexagonal spinner with 6 numbers in total. The probabilities for landing on specific numbers are:

  • The probability of landing on a 1 is \frac{1}{2}.
  • The probability of landing on a 2 is \frac{1}{6}.
  • The probability of landing on a 3 is \frac{1}{3}.

Write the numbers on the spinner.

First, convert each fraction to have a denominator of 6, as there are 6 numbers on this spinner:

  • 1 = \frac{1}{2} = \frac{3}{6}
  • 2 = \frac{1}{6}
  • 3 = \frac{1}{3} = \frac{2}{6}

Use the numerators to find how many times each number appears on the spinner:

  • 1 = \frac{3}{6} = 3 out of the 6 numbers
  • 2 = \frac{1}{6} = 1 out of the 6 numbers
  • 3 = \frac{2}{6} = 2 out of the 6 numbers

Complete the spinner.

Therefore, on the spinner, the numbers should be distributed based on these probabilities to make it a “fair” spinner. The sum of these probabilities is \frac{3}{6} + \frac{1}{6} + \frac{2}{6} = 1, confirming it as a fair spinner. Note that the placement of the numbers doesn’t matter.