### Video Transcript

Daniel spins two four-sided spinners, which are both numbered from one to four. Let π΄ be the event that he spins two numbers with the product of four, and let π΅ be the event that he spins two even numbers. Find the probability of π΄, giving your answer as a fraction in its simplest form. Find the probability of π΅, giving your answer as a fraction in its simplest form. Find the probability of the intersection of π΄ and π΅, giving your answer as a fraction in its simplest form. And finally, find the probability of π΄ union π΅, giving your answer as a fraction in its simplest form.

Beginning with part one, the probability of event π΄, for spinner one, Daniel could land on one, two, three, or four. If he spins a one on spinner one, then he could land on one, two, three, or four with his second spinner. The same thing is true if he spins a two, a three, or a four on the first spinner. What this means is that there are 16 total outcomes. He could spin a one and then a one; a one, two; a one, three; and so on. In order for event π΄ to occur, he needs to spin two numbers with the product of four, so we should multiply what he gets from spinner one and spinner two and see which of these equal four. If we multiply one times four, we get four. If we multiply two times two, we get four. And if we multiply four times one, we get four.

The combinations one, four; two, two; and four, one are the only ways for event π΄ to occur. This means there are three ways for event π΄ to occur out of the 16 possible outcomes, making the probability of event π΄ three sixteenths, which is already a fraction in its simplest form. And so we say the probability of event π΄ occurring is three sixteenths. Weβll follow a similar process to find the probability of event π΅. In order for event π΅ to occur, he has to spin two even numbers. If Daniel spins spinner one and gets a one, itβs already odd, and therefore he cannot get two even numbers. This is also true if he spins three on spinner one, so weβll focus on the cases if he spins two or four on spinner one.

If he spins a two on spinner one and a two or four on spinner two, then both are even numbers. So we have the combination two, two and two, four. If he spins a four on spinner one, to have two even numbers, he would have to spin a two or a four on spinner two, giving us the combination four, two and four, four. Out of the 16 total outcomes, four of them are the cases where Daniel spins two even numbers. The probability for spinning two even numbers is then four out of 16. Both four and 16 are divisible by four, which means the probability of event π΅ occurring in its simplest form is one-fourth.

For part three, weβre looking for the probability of the intersection of π΄ and π΅. If we think about a Venn diagram of events π΄ and π΅, the intersection of events π΄ and π΅ are the overlap. Itβs the probability that both are true at the same time. This means weβre looking for the places where Daniel spins values that have a product of four, and theyβre both even. Looking at the list for event π΄ and event π΅, this only occurs once in the event that both rolls are two. The only way for Daniel to roll two values with the product of four and two even numbers is that both spinners land on two. That is one of the outcomes out of a total of 16. And that means the probability that both π΄ and π΅ are true, the probability of the intersection of π΄ and π΅, is one sixteenth, which is of course in its simplest form.

Finally, we want to find the probability of π΄ union π΅. Sometimes we say this as the probability of π΄ or π΅. If we think about events π΄ and π΅ on a Venn diagram, the probability of π΄ union π΅ is all of the probabilities for π΄ or π΅. If we look at our list, we have the three cases where π΄ occurs and then we have three additional cases where π΅ occurs. We wanna be careful not to count the case two, two twice. Even though there were three ways for π΄ to occur and four ways for π΅ to occur, there are only six ways for π΄ or π΅ to occur. And thatβs because one of the values is shared between the two lists, which is where we get the formula that the probability of π΄ union π΅ is equal to the probability of π΄ plus the probability of π΅ minus the probability of the intersection of π΄ and π΅.

And so we can show that three sixteenths plus four sixteenths minus one sixteenth equals six sixteenths. And we can reduce this as both six and 16 are divisible by two. The probability of π΄ union π΅ is three-eighths.