### Video Transcript

The probability that it rains on a
given day is 0.6. If it rains, the probability that a
group of friends play football is 0.2. If it does not rain, the
probability that they play football rises to 0.8. Work out the probability that it
rains on a given day and the friends play football. Work out the probability that it
does not rain on a given day and the friends play football. What is the probability that the
friends will play football on a given day?

There are three parts to this
question. They all involve conditional
probability and the dependent events whether it rains and whether a group of friends
play football. One way to represent the
information from the question is using a tree diagram. We will now clear some space to do
this first.

We will begin by letting π
be the
event that it rains. We are told that the probability
that it rains on any given day is 0.6. We know that the complement of any
event π΄, which is written π΄ prime or π΄ bar, has a probability that is equal to
one minus the probability of π΄. This means that the probability it
does not rain in this question is one minus 0.6. This is equal to 0.4 and can be
added to the tree diagram as shown.

If we let the event that the group
of friends play football be πΉ, there are four possible scenarios: firstly, that it
rains and the friends play football; secondly, that it rains and the friends do not
play football; thirdly, that it does not rain and the friends play football; and
finally, that it does not rain and the friends do not play football. We are told that if it rains, the
probability that the friends play football is 0.2. This is an example of conditional
probability, the probability that the friends play football given that it rains. We can then add 0.2 to our tree
diagram.

Once again, since the probabilities
on each pair of branches sum to one, the probability of the complement of this is
0.8. The probability that the friends do
not play football given that it rains is 0.8. We can repeat this for the bottom
half of our tree diagram. We are told in the question that if
it does not rain, the probability that the friends play football is 0.8. The conditional probability that
the friends play football given that it does not rain is 0.8.

Letβs now return to the three
specific questions we were asked. Firstly, we were asked to work out
the probability that it rains on a given day and the friends play football. As we want both events to occur,
this is the intersection of the two events. We recall that given two events π΄
and π΅, the probability of π΄ intersection π΅ is equal to the probability of π΅
given π΄ multiplied by the probability of π΄. In this question, the probability
that it rains and the friends play football is equal to the probability that the
friends play football given that it rains multiplied by the probability it
rains. We need to multiply the
probabilities 0.2 and 0.6. This is equal to 0.12.

Letβs now consider the second part
of our question. This asked us to work out the
probability that it does not rain on a given day and the friends play football. This corresponds to the pink path
on our tree diagram. The probability that it does not
rain and the friends play football is equal to the probability they play football
given that it does not rain multiplied by the probability it does not rain. We need to multiply 0.8 and
0.4. This is equal to 0.32. The probability that it does not
rain on a given day and the friends play football is 0.32.

The final part of our question
asked us to calculate the probability that the friends play football on a given
day. This can occur in one of two ways:
either it rains and they play football or it does not rain and they play
football. We need to find the union of these
two events. From the tree diagram, this
involves finding the sum of the probabilities. We need to add 0.12 and 0.32. This is equal to 0.44. We can therefore conclude that the
probability that the friends play football on a given day is 0.44.

It is worth noting that the sum of
the probabilities for every possible outcome combined is equal to one. In this case, the four
probabilities 0.12, 0.48, 0.32, and 0.08 sum to one.