Video Transcript
Benjamin spins two spinners. The first has six equal sectors numbered from one to six. And the second has four equal sectors numbered from one to four. He draws a two-way table to represent the sample space, as shown in the figure. Work out the probability that at least one of the spinners lands on a two. Work out the probability that the sum of the numbers is even. Work out the probability that at least one of the spinners lands on a two and the sum of the numbers is even. Work out the probability that the sum of the numbers is even given at least one of the spinners lands on a two.
So, let’s begin by having a look at the two-way table. The values along the columns represent the spinner that has six equal sectors from one to six, and the values down the side represent the spinner that has the four equal sectors. This two-way table represents the outcomes when the values on each of the two spinners are added together. For example, this value of two comes from adding the one on each of the two spinners. The second value of three represents a one on one spinner and a two on the other spinner, and so on.
Importantly, we can also see from this two-way table that there must be 24 different outcomes. This will be helpful when it comes to answering the questions. So, let’s have a look at the first question. Here, we need to work out the probability that at least one of the spinners lands on a two. This would be all the values in this column and all the values in this row. This value of four in the overlap represents when we have a two on both spinners. But that would still be counted in the probability that at least one of the spinners lands on a two. There is a total of nine different outcomes. And we know that there’s 24 outcomes altogether. We can simplify by dividing the top and the bottom of this fraction by three to give us an answer of three-eighths.
In the second question, we’re asked to find the probability that the sum of the numbers is even. So, here we can use the sums given in the sample space diagram to help us. If we highlight all the sums that are even, we’ll have 12 of these out of a total of 24 different outcomes. 12 over 24, of course, simplifies to the fraction of one-half, which is the answer for the second question.
In the third question, we need to identify the probability that at least one of the spinners lands on a two and the sum of the numbers is even. So, just like in the first question, we’re looking for the values in this column and in this row with the spinners on two. But we’re just looking for those values which have an even sum, which means that it’s just the sums of four, six, eight, and six. And that’s a total of four different outcomes out of 24. We can simplify this fraction four twenty-fourths to give us the fraction of one-sixth.
In the final question, we’re identifying a conditional probability, as we have the probability of an event given another event. It will be helpful to remember the rule that to find the probability of 𝐴 given 𝐵, it’s equal to the probability of 𝐴 intersection 𝐵 divided by the probability of 𝐵.
The probability of 𝐴 intersection 𝐵 is like the probability of 𝐴 and 𝐵. As we need to find the probability of an even sum given at least one of the spinners lands on a two, we’ll need to divide the probability of getting an even sum and at least one on a two by the probability of getting at least one spinner landing on a two. That means we’ll therefore be dividing one-sixth, that was the answer in our third question, by the value of three-eighths, which was what we got in our first question.
In order to work out one-sixth divided by three-eighths, we’ll need to remember that when we’re dividing fractions, it’s equivalent to multiplying by the reciprocal of the second fraction. We, therefore, calculate one-sixth multiplied by eight over three. We can simplify the values before we multiply the numerators and denominators, giving us a fraction of four-ninths. And so, that’s the answer for the fourth and final question. The probability that the sum of the numbers is even given at least one of the spinners lands on a two is four-ninths.
