Video Transcript
In this video, we’re gonna look at
events, which are not necessarily mutually exclusive, and examine the probability of
one or other or both of them occurring. This can take a little bit of
thinking about.
Let’s think about a class of thirty
children. Let’s say that fifteen of them
regularly play a musical instrument, and twenty of them play some sort of a sport
for a team. Now fifteen plus twenty is
thirty-five, which is more than thirty. So some of the children must play
an instrument and play a sport. Now if I tell you that twelve of
them play both, an instrument and a sport, we can create a diagram which neatly
breaks down our numbers. So we’re gonna do a Venn
diagram. Now before we start, let’s let 𝐼
be the number of children who play an instrument, and 𝑆 be the number of children
who play a sport. So we’ve got the total class here
of thirty students, so thirty students are represented on the diagram. And we’ve got those who play an
instrument in the left circle, and those who play a sport in the right-hand
circle. Right. Now we know that twelve students
play both. So in this intersection area in the
middle, there are twelve students who are within the instrument circle and also
within the sports circle. Now if fifteen of them play an
instrument in total, but twelve of those also play a sport, because that’s what we
said here, then fifteen take away twelve leaves three students who purely play an
instrument but don’t play a sport. And we know that twenty play a
sport in total. Twelve of those also play an
instrument, so twenty take away twelve leaves eight of them who play sport but don’t
play an instrument.
So the class is made up of thirty
students, three purely play an instrument, twelve play both an instrument and a
sport, and eight play a sport. So if we take those numbers away
from thirty, that gives us an answer of seven. It tells us that seven of the
students don’t play an instrument or a sport. Now we can use our counting cases
technique to help us work out the various probabilities if we pick a child at
random. So if we do this, each child is
equally likely to be picked. So the probability that they play
an instrument, 𝑃𝐼, is simply equal to — well there are three who just play an
instrument and there are twelve who also play a sport, so if we add those together,
that’s fifteen in total who play an instrument out of the thirty people in the class
— so the probability they play an instrument is fifteen out of thirty, fifteen
thirtieths. For playing a sport, again, we’ve
got twelve who play a sport and an instrument and eight who just play a sport, so
add those two together and that will tell us what proportion of the thirty play
sport. And that’s kind of reassuring cause
that is in fact what we were told at the beginning of the question. And then finally, we did the
calculations already for this one, the probability that they don’t play an
instrument and they don’t play a sport, well there was seven in that category out of
the thirty students in the class, so the probability they don’t play an instrument
or a sport is seven over thirty.
Now let’s introduce a little bit of
notation. 𝐼 intersection 𝑆, this symbol
here, it’s the group of students in the intersection. So they’re in both groups, in this
group here. So these twelve students who are in
the instrument group and also in the sport group. So provided I’m picking a student
at random from the class, so they are all equally likely to be picked, the
probability of 𝐼 intersection 𝑆, the probability that they played both an
instrument and a sport, is twelve out of thirty because there are twelve in that
category out of thirty people in total in the class.
Now this symbol here means 𝐼 union
𝑆 and it’s a count of the twenty-three students who either play an instrument, or
play a sport, or play both. So they’re in the 𝐼, or in the 𝑆
categories, or indeed both. Now notice that we have to be quite
careful about how we work this out. If I just added together the
fifteen who played an instrument to the twenty who played sport, I’d have counted
these twelve here in both groups. So I’d have counted them twice and
I’d get an answer that’s too large by twelve. So again, so long as I’m picking
students at random and they’re all equally likely to be picked, to calculate 𝐼
union 𝑆, I just have to add up all of these three numbers here to get the total of
students who are in those three- in those two categories.
Now one last little bit of
notation, the compliment or dash. This here, you could pronounce it
𝐼 dash, or 𝐼 compliment; it means not 𝐼. So that’s the students who don’t
play an instrument. So they could be one of these eight
who don’t play an instrument but do play a sport, or they could be one of these
seven who don’t play an instrument and don’t play a sport either. So since students either do play an
instrument or they don’t play an instrument, then we can say this. It’s absolutely certain, a
probability of one, that they either do or don’t play an instrument. So if we add together the
probability that they do play an instrument to the probability that they don’t play
an instrument, they must sum to one. So if you wanted to work out the
probability that a student doesn’t play an instrument, we can just rearrange that
formula 𝑃𝐼 dash equals one minus 𝑃𝐼, the probability that they do play an
instrument. So remembering back, we worked out
the probability they do play an instrument, was fifteen out of thirty. In fact, we were given that in the
question. So the probability they don’t play
an instrument is one minus that, which also comes out to be fifteen over thirty.
And applying the same principle to
those who play or don’t play sport, the probability that somebody doesn’t play sport
is one minus the probability that they do play sport. And we were told that the
probability they do play sport is twenty over thirty because twenty out of the
thirty students play sport. So one take away twenty over thirty
gives us an answer of ten thirtieths or one-third. So the probability they don’t play
sport is ten thirtieths.
Okay. Just one more thing then. Let’s try a few of those little
techniques together. What do we think this means, 𝐼
union 𝑆 all dashed? Well 𝐼 union 𝑆 is the int- is the
complete set of students who play a sport or an instrument or both. So that’s all of these people
here. So that’s 𝐼 union 𝑆. Now we want dash, the complement of
that. So it’s all the people outside that
group. So that’s these seven students over
here. So the Venn diagram actually helps
us to really visualize what that means. So let’s just kind of formalise
that a bit. The probability then, if all the
students are picked at random, of 𝐼 union 𝑆 dashed, all of those not in that
group, is one minus the probability of 𝐼 union 𝑆. And since there are twenty-three
students who are in 𝐼 union 𝑆, who either play an instrument or a sport or both,
the probability then of picking one of those would be twenty-three thirtieths. So the probability of picking
someone who doesn’t play an instrument or a sport is one minus twenty-three
thirtieths, which is seven thirtieths. And again, that’s great. That’s the answer that we could
have got just by visual inspection. But we’ve now got different ways
that we can work out the same thing. So we can check our answers and we
can approach things in different ways when we’ve got different questions.
So let’s just summarise what we’ve
learnt so far. And now this is the result really
that we trying to find. The probability of 𝐴 union 𝐵 is
equal to the probability of 𝐴 plus the probability of 𝐵 minus the probability of
𝐴 intersection 𝐵. So how does that work? Well the probability of 𝐴 is this
region here, so these all the ones that are in the 𝐴 group. And the probability of 𝐵 is all
these people in here. Now what we can see is in this-this
intersection region here, we’ve actually counted those people twice. We’ve counted them once amongst the
𝐵 group and also once amongst the 𝐴 group. So that’s why we’re subtracting 𝐴
intersection 𝐵 from that total because they’re the- there we’re just basically
removing the ones that we’ve double counted to make that calculation. So now that we understand that, we
can sometimes, we don’t have to draw a Venn diagram for every question, we can just
use that formula straightaway to do some of these calculations for some of our
questions. So let’s have a look at o- a case
where we do that.
So the question is, find the
probability of getting a number which is odd or prime when rolling a fair
six-sided dice. So if it’s odd or prime, this
tells us we’re trying to find the union of odd and prime. So it could be odd or prime or
perhaps even both, when we roll this dice.
So let’s just right down the
definitions of our events then. We’re defining an odd event as
getting an odd number, so that’s one or three or five. All of which are equally likely
when we roll a dice. Or the event prime is when we
get a prime number, so the prime numbers between one and six are two and three
and five. And the question is asking us
for the probability of getting an odd or a prime, which is the same as odd union
prime. And because we’ve got equally
likely outcomes for rolling a dice, it’s a fair dice, we can say that the
probability of getting an odd number, we can use our counting technique there,
so three ways out of six in total. And the same for prime numbers,
there are three ways of getting a prime number out of the six possible results
that you can get when you roll a fair six-sided dice.
So to work out the probability
of odd union prime, we’re gonna be adding together the probability of odd plus
the probability of prime. And then we’re taking away the
intersection of those two, the probability of the intersection of those two. So the next thing we need to
work out, is what’s the probability of getting a number which is both odd and
prime. Well if we look at those two
lists, one is odd but it isn’t prime, two is prime but it isn’t odd. So three and five are the only
two elements in that set. So there are two ways of
getting a number which is odd and prime out of the six total possible
results. Then using the standard result,
the probability of odd union prime is the probability of odd plus the
probability of prime minus the probability of the intersection of odd and
prime. And so let’s just plug in those
numbers that we found, so that’s three-sixths plus three-sixths take away
two-sixths, which is four-sixths.
And just a quick mental check,
that makes sense, doesn’t it? Because four of the six numbers from one to six
are either odd or prime or both. So that’s one, two, three, and
five.
Okay. So here is a question then.
In a survey of people buying
gelato at a cafe, sixty percent said they liked strawberry and forty percent
said they liked pistachio. Analysis showed that seventy
percent liked either strawberry or pistachio or both. What proportion liked both, and
what proportion just liked strawberry?
So first up, let’s define
our-our events. So let 𝑆 be the event that the
person liked strawberry, let 𝑃 be the event that the person liked
pistachio. So just filling in the numbers
from the question, the probability that they liked strawberry was sixty percent,
that’s nought point six, the probability that they liked pistachio was forty
percent, so that’s nought point four, and the probability that they liked either
or both was a nought point seven, so seventy percent, so this probability there
is nought point seven.
Now our general result tells us
that the union- the probability of 𝑆 union 𝑃, so the probability that they
liked either strawberry or pistachio or both, is equal to the probability they
liked strawberry plus the probability they liked pistachio minus the
intersection of those two cause remember, we’d have double-counted those in that
union. So we need to take those back
out again. Now what we’re looking for is,
what proportion liked both. So that’s this proportion
here. The probability that it’s 𝑆
intersection 𝑃, that they-they liked both strawberry and pistachio. So we can rearrange this
equation here, so adding the probability of 𝑆 intersection 𝑃 to both sides and
subtracting the probability of 𝑆 union 𝑃 t- from both sides. And now we can substitute in
the values that we’ve got. So that’s nought point six plus
nought point four take away nought point seven, which gives us an answer of
nought point three. In other words, thirty percent
of the people surveyed liked both strawberry and pistachio ice cream.
Now the last part of the
question, what proportion just liked strawberry. In other words, didn’t like
pistachio. Well we’ve just worked out that
this intersection area here is nought point three. So the question told us that
the probability that we would like strawberry is nought point six. So nought point three are
accounted for here. So we’ve got to take that away
from the nought point six to work out what’s left in here. So that region is, let’s call
it 𝑆 intersection 𝑃 dash, so they- it’s the intersection of them liking
strawberry and not liking pistachio. And that is the-the probability
they liked strawberry take away the probability that they liked both. So that’s nought point six take
away nought point three, which again is nought point three. So there’s that final result,
thirty percent then, nought point three just like strawberry i- gelato. So nought point six minus
nought point three is nought point three. So thirty percent of people
surveyed just liked strawberry but didn’t like pistachio. So we can write that as our
answer as well.
So top tips here then were, one
defining our events, two using the standard result that the union of two events
is equal to probability of 𝑆 plus probability of 𝑃 minus the intersection of
those two events there. Also, this technique of just
using the Venn diagram to help us to gather our thoughts is really useful. And make sure that you make
your answer nice and clear at the end.
Okay. So if there’s one thing that you
remember from this video, let this be it. The probability of 𝐴 union 𝐵 is
equal to the probability of 𝐴 plus the probability of 𝐵 minus the bit that we
double counted, the intersection of those two areas there.
