Video Transcript
In an experiment, Ethan is going to spin a fair three-sided spinner and a fair four-sided spinner. He draws a two-way table to show all of the possible outcomes. In his experiment, he wants to look at two events: spinning two numbers whose sum is prime, 𝐴, and spinning at least one three, 𝐵.
There are six parts to this question which we will look at in turn. Before doing so, let’s consider what we mean by event 𝐴 and event 𝐵. We recall that prime numbers have exactly two factors, the number one and the number itself. The first four prime numbers are two, three, five, and seven. Event 𝐴 is when the sum of our two numbers is prime. These are highlighted in orange on the two-way table. One plus one is equal to two, which is a prime number. One plus two and two plus one equals three, another prime number. Three plus two, two plus three, and one plus four all equal five, and three plus four is equal to seven.
Event 𝐵 occurs when Ethan spins at least one three. This consists of all the events in the bottom row — three, one; three, two; three, three; and three, four — and all the events in the third column — one, three; two, three; and three, three. Note that we will not need to count the event three, three twice.
The first part of our question asks us to find the probability of event 𝐴.
As seven of the 12 outcomes, those in orange, have a sum that is prime, the probability of event 𝐴 is seven out of 12 or seven twelfths.
The second part of our question asks us to find the probability of event 𝐵.
Six out of the 12 outcomes, those circled in pink, have at least one, three. This means that the probability of event 𝐵 is six out of 12. As both the numerator and denominator are divisible by six, this simplifies to one out of two or one-half. We will now clear some space and consider the remainder of this question.
The third part of the question asks us to find the probability of 𝐴 given 𝐵.
This is known as conditional probability. We saw that there were six outcomes in event 𝐵. Of these, three are also in event 𝐴. These are the elements in the table that are circled both orange and pink: two, three; three, two; and three, four. There are six ways that Ethan can spin at least one three, and three of these have a sum that is prime. By dividing the numerator and denominator by three, the fraction three-sixths simplifies to one-half. The probability of 𝐴 given 𝐵 is one-half.
The fourth part of our question asks us to find the probability of 𝐵 given 𝐴.
We know that seven outcomes occur in event 𝐴. These are the elements circled in orange, and of these, three of them are also in event 𝐵. Of the seven outcomes where the two spinners are prime, three of them have at least one three, the elements two, three; three, two; and three, four. The probability of 𝐵 given 𝐴 is therefore equal to three-sevenths.
Next, we are asked to find the probability of 𝐴 intersection 𝐵.
As already mentioned, there are three outcomes where both event 𝐴 and event 𝐵 occur: two, three; three, two; and three, four. The probability of 𝐴 intersection 𝐵 is therefore equal to three out of 12 or three twelfths. And by dividing the numerator and denominator by three, this simplifies to one-quarter. The last part of our question asks the following.
Is it true that the probability of 𝐴 multiplied by the probability of 𝐵 given 𝐴 is equal to the probability of 𝐴 intersection 𝐵 and the probability of 𝐵 multiplied by the probability of 𝐴 given 𝐵 is also equal to the probability of 𝐴 intersection 𝐵?
After clearing some space, let’s consider the first statement. The probability of 𝐴 multiplied by the probability of 𝐵 given 𝐴 is equal to the probability of 𝐴 intersection 𝐵. On the left-hand side, we have seven twelfths multiplied by three-sevenths. By firstly cross canceling, we see that this is equal to one-quarter, which is the value of the probability of 𝐴 intersection 𝐵. The first statement is therefore correct.
The second statement said that the probability of 𝐵 multiplied by the probability of 𝐴 given 𝐵 is also equal to the probability of 𝐴 intersection 𝐵. This time, the left-hand side is equal to one-half multiplied by one-half, which once again is equal to one-quarter. We can therefore conclude that it is true that the probability of 𝐴 multiplied by the probability of 𝐵 given 𝐴 is equal to the probability of 𝐴 intersection 𝐵 and the probability of 𝐵 multiplied by the probability of 𝐴 given 𝐵 is also equal to the probability of 𝐴 intersection 𝐵. This is actually a general rule that can always be used when dealing with conditional probability.
The six answers to this question are seven twelfths, one-half, one-half, three-sevenths, one-quarter, and yes.
