Lesson Video: Calculating Probabilities with Equally Likely Outcomes Mathematics • 7th Grade

Express the number of cases representing an event as a proportion of all cases to find its probability if individual outcomes are all equally likely.

06:06

Video Transcript

If all of the possible outcomes from an experiment are equally likely to occur, then calculating the probability of an event becomes a simple matter of counting up the number of outcomes that make up the event, and express that as a fraction of the total number of possible outcomes.

So, for example, if we roll a fair six-sided dice, there are six possible outcomes. We can get a one, two, three, four, five, or six and they’re all equally likely. Now one of the outcomes is a three. So there’s one way to get a three. So the probability of getting a three is this one divided by the total number of outcomes, six. So the probability of getting a three is one over six.

So when all of the outcomes are equally likely, to work out the probability of an event, we get the number of ways to get the result we’re looking for and divide that by the total number of outcomes that we have altogether. But remember, our method of counting outcomes to work out probabilities, only works when the outcomes are all equally likely. So let’s look at an example where that isn’t the case.

If we buy a lottery ticket, there are two possible outcomes. We can win, or we can lose. But most lotteries, those two don’t have the same probability. We’re much more likely to lose than we are to win. So they’re not equally likely outcomes. So although we have two outcomes in total, and there’s one way to win. The probability of winning is not just one divided by two. It’s not a half because those two outcomes were not equally likely. Remember, this was much more likely to occur than this. So we can’t just do the counting. So, as always, something to work, to look out for, are the outcomes equally likely?

Okay. Now let’s go back to our dice example, where the outcomes were all equally likely. Let’s work out the probability of the event of getting a factor of twelve, when we roll our dice. So the first thing we need to do, is see which of the results would go into event 𝐸, so which of those numbers are factors of twelve. Well one is, two is, three is, four is, five isn’t, and six is. So that’s five equally likely outcomes making up event 𝐸. So the probability that event 𝐸 occurs, that we get a factor of twelve when we roll our dice, well there’re five ways of that happening, so that’s five, out of the six possible outcomes that we had in total, so that’s five over six. So with equally likely outcomes, it’s just a matter of counting up the cases. There are five ways of getting a factor of twelve out of six possible equally likely ways of things turning out in the end.

Now this works in slightly more complex situations too. So let’s say we roll two fair dice and add the scores together. We can put the results in a table like this. We’ve got thirty-six possible equally likely outcomes. So, for example, we could get a one on the first dice and a one on the second dice, making a total of two. We could get a one on the first dice and a two on the second dice, making a total of three, and so on. So let’s ask the question, what is the probability of getting a result of nine? Well there are thirty-six possible equally likely outcomes and one, two, three, four of them result in a nine. So the probability of getting a nine is four out of thirty-six, four ways of getting a nine out of thirty-six possible equally likely outcomes in total.

Now let’s ask a question, what’s the probability of getting an odd result? Well we can see that eighteen out of the thirty-six equally likely outcomes give odd results, odd-numbered results. So the probability of getting an odd number is eighteen over thirty-six. Now we could simplify that to a half, but we don’t have to. In probability, quite happy to leave that eighteen over thirty-six. Again, it’s kind of more informative than a half in some ways because there are eighteen ways of getting what we’re looking for, out of the thirty-six total possible ways that there are of getting results.

Okay. Now let’s consider an experiment where we roll two fair dice, and we multiply their scores together. Again, we’ve got thirty-six possible equally likely outcomes. We could get a one on the first dice and a one on the second dice which would — one times one gives us a result of one. Or, we could get a one on the first dice and a two on the second dice, and one times two gives us a result of two, and so on, and so on, for all thirty-six different examples. So when we write out all the results in a sample space, in a table, each cell is equally likely to occur. And again, we can just count up the ways of results happening.

So let’s ask the same questions again, what’s the probability of getting a result of nine? Well there are thirty-six possible outcomes, all equally likely. But only one of them here, if we get a three and a three, generates a result of nine. So there’s one way out of thirty-six of getting a nine.

And, so what’s the probability of getting an odd-numbered result? Well now that we’re multiplying the numbers together on the dice, there’s only nine ways of getting an odd result, so nine out of thirty-six. The probability of getting an odd number result in this situation is a lot lower than it was last time.

So to sum all that up, if all of our outcomes are equally likely in an experiment, if we’ve got a sample space of 𝑆 and 𝐸 is an event that we’re looking for, to work out the probability of event 𝐸 occurring, we just count up how many ways are there of getting 𝐸 versus how many ways were there in the-in the sample space altogether. And then just express that as a fraction.

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