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
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,
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.