Video Transcript
In this video, we will learn how to
find the probability of simple events. So let’s begin by thinking about
what probability actually is. Probability is the likelihood or
chance of an event occurring. We might even see this given as the
likelihood of an outcome occurring. When we first started learning
probability, and even in everyday life, we use words to describe the probability of
something happening. For example, we might say that it’s
impossible for an event to occur. Or if it’s definitely going to
happen, then we might say that it’s certain to happen. We can also use words like unlikely
and likely.
When we’re discussing probability
in mathematics, it’s good to assign numerical values to probability. An impossible event has a
probability of zero, and an event that’s certain to happen has a probability of
one. We can describe probabilities
either as decimals or as fractions.
In this video, we focus on the
probability of simple events. That’s an event with a single
outcome. Let’s say, for example, that we’re
going to toss a coin. One side of the coin has heads, and
the other side is tails. If we wanted to work out the
probability of getting tails, we could write it like this: the capital P and then
tails in parentheses. To work out the numerical value, we
could say that one out of two outcomes is tails. So we could write it like this: the
fraction one-half. Alternatively, if we were throwing
a die that has the numbers one through six on it, then if we wanted to work out the
probability of rolling a three, we know that one out of six possible numbers will be
three. And so the probability of rolling a
three would be one-sixth.
In general then, we can say that
for a simple event, the probability of this event is equal to the number of possible
outcomes over the total number of outcomes. Remember that when we were looking
at the coin, the probability of tails is the one outcome that’s tails out of the two
in total, that is, heads or tails. We will now look at a few questions
and see how we can apply this formula.
A class has 18 boys and nine
girls. What is the probability that a
randomly selected student is a girl?
In this question, we’re told that
there’s a class made up of 18 boys and nine girls. We need to find the probability
that if we were just to pick one of these students, we would pick a girl. A really common misconception is to
think that if we just picked one of these students, we’d either get a boy or a
girl. We might therefore think that the
probability is 50/50 or a half. However, this is incorrect. After all, if we think about it,
there’s more boys than girls. Therefore, we’re more likely to
pick a boy than a girl.
When we have an event like this,
where we’re picking something and there’s just one outcome, in this case boy or
girl, then we can apply the rule that the probability of this event is equal to the
number of possible outcomes over the total number of outcomes. So in this question, because we
want to find the probability of picking a girl, then the number of possible outcomes
in this case would be the number of girls. The total number of outcomes would
be the total number of students.
Once we’ve written the equation in
the context of the question we’re working with, then we just need to fill in any
values we’ve been given. We’re told that there are nine
girls. But be careful because the total
number of students is not 18 because that’s the number of boys. To find the total number of
students, we add together the boys and the girls, giving us 27. Of course, it’s always good to
simplify a fraction when we can. And nine over 27 simplifies to
one-third. And so we found that the
probability of picking a girl is one-third.
Let’s have a look at another
question.
What is the probability of the
pointer landing on an even number when the given spinner is spun?
When we have a question like this,
we can assume that the spinner is fair. And by that we mean it’s not biased
or weighted, for example, so that the pointer will land more frequently on one
section than another. We can see that all the sections
are the same size. In particular, the angle made at
the center of this spinner in each sector will be the same.
Because there will just be one
outcome any time the spinner is spun, then we can use a key probability formula. The probability of an event is
equal to the number of possible outcomes over the total number of outcomes. So what event are we interested in
here? Well, it’s the probability of the
pointer landing on an even number. And we can write this as P and even
in parentheses. Then the number of possible
outcomes is the number of even values on the spinner. And the total number of outcomes is
the total number of values on the spinner.
So let’s work out the even and the
odd values on the spinner. Well, just 12 and 14 are even, so
that’s two even values. And then there’s eight different
values on the spinner. Notice that we’ve also included the
even numbers as they’re still part of the total number of values. Two over eight simplifies to
one-quarter. So the answer is that the
probability of the pointer on this spinner landing on an even number is
one-quarter.
Let’s take a look at another
question.
A deck of cards contains cards
numbered from one to 81. If a card is picked at random, what
is the probability of picking a card that is divisible by five?
So let’s consider this deck of
cards. The cards are numbered from one to
81. Because we’re considering the
probability of picking a card or a particular type of card, we can use this
equation. The probability of an event is
equal to the number of possible outcomes over the total number of outcomes. So in this question, the
probability of picking a card which is divisible by five is equal to the number of
card values which are divisible by five over the total number of cards.
Let’s remember what it would mean
for a value to be divisible by five. Any value that’s divisible by five
means that we can divide that value by five and get an integer solution. An alternative way of thinking
about it would be the values which are in the five times tables. We can list all the values here
which are divisible by five. The first one would be five. The second one would be 10. And we can continue until we get to
the final one of 80. We can’t go any higher because the
cards only go up to 81. When we count up these values,
there are 16. That means that the number of card
values that are divisible by five is 16, and the total number of cards must be
81. We can’t simplify this fraction any
further. So the probability of picking a
card that is divisible by five is 16 over 81.
In the next question, we’ll see how
we can use the same formula, only this time we’re not working like the probability,
but instead the number of outcomes of a specific event.
There are 28 people in a
meeting. The probability that a person
chosen at random is a man is one-half. Calculate the number of women in
the meeting.
Here, we’re told that there are 28
people in the meeting. If we pick a person at random, then
the probability of selecting a man is one-half. We can use these two pieces of
information to help us work out the number of women there must be in the
meeting. We can use the equation that the
probability of an event is equal to the number of possible outcomes over the total
number of outcomes. One of the ways we can approach
this question is because we know the probability of selecting a man, we can work out
the number of men in the meeting. The probability of picking a man is
equal to the number of men over the total number of people.
We’re given that the probability of
picking a man is one-half. The number of men is what we want
to find out, and the total number of people is the number of people in the
meeting. That’s 28. In order to simplify this, we can
multiply both sides by 28. And since a half multiplied by 28
is 14, we now know that the number of men is 14. So if there’s 28 people in total in
the room, either men or women, and 14 of those are men, then subtracting that would
give us 14. So we can give the answer then that
there must be 14 women in the meeting.
In the final question, we’ll see
how we can use the formula along with the given information to work out a total
number of outcomes.
A bag contains 24 white balls and
an unknown number of red balls. The probability of choosing at
random a red ball is seven over 31. How many balls are there in the
bag?
So let’s say that we’ve got this
bag which has 24 white balls. It also has an unknown number of
red balls. So there could be one or two or
three or even more than 24. We don’t know. What we are told, however, is that
the probability of choosing a red ball is seven over 31. We can answer this question in one
of two different ways, either by finding the number of white balls first or by
finding the number of red balls first.
In order to find the number of
white balls first, we need to remember that in any situation, the probabilities will
sum to one. Because we only have white balls
and red balls in the bag, then we can say that the probability of getting a white
plus the probability of getting a red must be equal to one. We can rearrange this to give us
that the probability of getting a white is equal to one minus the probability of
getting a red ball. As we’re told that the probability
of getting a red is seven over 31, then we need to work out one minus seven over
31. Since one can be written as 31 over
31, then we evaluate this as 24 over 31. So now we know that the probability
of picking a white is 24 over 31.
Because this is a simple event,
that is, an event with a single outcome, then we can use the fact that the
probability of an event is equal to the number of possible outcomes over the total
number of outcomes. We can use the information that
we’ve got about the white balls. We can say that the probability of
picking a white is equal to the number of white balls over the total number of
balls. So filling in the information, the
probability of a white ball is 24 over 31. And we were told in the question
that there are 24 white balls. And we need to work out the total
number of balls. So now we have this equation with
two fractions that are equivalent to each other. However, since the numerators are
equal to each other, they’re both 24, then the denominators must also be equal to
each other, which means that the total number of balls in the bag must be 31.
Before we finish with this
question, let’s have a look at the alternative method of finding the number of red
balls. We can keep the same probability
equation, only this time we’ll fill in the information about the red balls. We were told that the probability
of a red ball is seven over 31. We don’t know the number of red
balls, but we can use a little bit of algebra. And let’s define the number of red
balls with the variable 𝑥. The total number of balls then will
be the number of red balls, that’s 𝑥, plus the number of white balls, that’s
24.
We could then solve this by
starting with the cross product. So we’d have seven multiplied by 𝑥
plus 24 is equal to 31𝑥. Distributing the seven across the
parentheses would give us seven times 𝑥, and seven times 24 is 168. Subtracting seven 𝑥 from both
sides, we’d have 168 is equal to 24𝑥. Then dividing both sides by 24,
we’d have that seven is equal to 𝑥. Since we defined 𝑥 to be the
number of red balls, then we’ve worked out that the number of red balls in this bag
is seven. We didn’t just want to find the
number of red balls, however; we wanted to find the total. There are 24 white balls and seven
red balls. So that would give us 31 in total,
which confirms the original answer.
We can now summarize the key points
of this video. We began by recalling that the
probability of an event is the likelihood of it occurring. All of the questions that we looked
at were simple events. Those are ones that have a single
outcome, for example, rolling a die or selecting a ball from a bag. If, for example, instead we had
picked two balls from a bag, then that would not have been a simple event. Finally, we saw and applied the
formula that for a simple event, the probability of that event is equal to the
number of possible outcomes over the total number of outcomes.
