Three events for an ordinary fair dice are listed. A) Roll a one, B) roll a prime number, C) not rolling a six. Draw and label arrows on the number line to show the probabilities of events B and C occurring.
The probability of an event occurring can be written as a fraction: the number of successful outcomes over the number of possible outcomes. An ordinary fair dice has six sides. Therefore, there are six outcomes: the numbers one, two, three, four, five, and six.
Let’s firstly consider event A rolling a one. There is one number one on a dice. Therefore, the probability of rolling a one on a fair dice is one-sixth or one out of six.
Event B wants us to calculate the probability of rolling a prime number. Prime numbers have exactly two factors: the number one and the number itself. There are three numbers on a dice that are prime. They are two, three, and five, as all three of these numbers have exactly two factors. Therefore, the probability of rolling a prime number is three out of six or three-sixths.
This fraction can be simplified as the numerator and denominator are both divisible by three. Remember, with fractions, whatever you do to the top you must do to the bottom. Three divided by three is equal to one, and six divided by three is equal to two. Therefore, the probability of rolling a prime number is equal to one-half.
Finally, we have event C the probability of not rolling a six. The numbers one, two, three, four, and five are not six. This means that the probability of not rolling a six on a fair dice is five out of six or five-sixths.
We now need to label these probabilities on the number line. Event A has already been labelled. Therefore, the first hash mark corresponds to one-sixth, as the probability of rolling a one was one-sixth. This means that each of the other hash marks must represent two-sixths, three-sixths, four-sixths, and five-sixths. We already know that three-sixths is the same as one-half. In the same way, we can simplify two-sixths and four-sixths to one-third and two-thirds, respectively. In both cases, we divide the numerators and denominators by two.
The probability of rolling a prime number was three out of six or one-half. Therefore, B must point to three-sixths. The probability of not rolling a six was five out of six. Therefore, C must point of five-sixths on the number line. We have drawn and labelled arrows showing the probabilities of events B and C occurring.