### Video Transcript

In this video, we will learn how to
use tree diagrams to calculate conditional probabilities. When working with conditional
probabilities, it is helpful to use a tree diagram to illustrate the probability of
the different outcomes. To help understand this, letβs
first recall the formula for conditional probability.

The probability that an event π΅
occurs given that event π΄ has already occurred is written the probability of π΅
given π΄ is equal to the probability of π΄ intersection π΅ divided by the
probability of π΄, where the probability of the intersection is the probability that
both π΄ and π΅ occur. By multiplying both sides by the
probability of π΄, this can be rewritten as the probability of π΄ intersection π΅ is
equal to the probability of π΅ given π΄ multiplied by the probability of π΄.

We will now consider how this can
be represented on a tree diagram. We can use the formula to determine
the intersection of two events. On a tree diagram, these can be
calculated by multiplying across branches, with the first branch representing the
probability of π΄ and the second branch representing the probability of π΅ given
that π΄ has occurred. Recalling that all probabilities
lie in the interval from zero to one such that the probability of π΄ is greater than
or equal to zero and less than or equal to one and that the probability of the
complement of an event written π΄ prime or π΄ bar is equal to one minus the
probability of π΄, we can then complete the top half of the tree diagram.

Extending this to the bottom half,
beginning with the probability of π΄ prime, the completed tree diagram is as
shown. By multiplying along our branches,
weβre able to calculate the probabilities of π΄ intersection π΅, π΄ intersection π΅
prime, π΄ prime intersection π΅, and π΄ prime intersection π΅ prime. Letβs now consider how we can use a
tree diagram of this type in context.

A bag contains three blue balls and
seven red balls. David selects two balls without
replacement and draws the following tree diagram. Given that the first ball is red,
find the value of π₯ that represents the probability that the second ball selected
is red.

We are told in the question that
the two balls are selected without replacement. This means that the color of the
second ball is dependent on the color of the first ball. As a result, we need to consider
what the outcome of the first event is when calculating the probability of the
second event. This is an example of conditional
probability. We are trying to find the
probability that the second ball is red given that the first one was red.

We are told initially that the bag
contains three blue balls and seven red balls. As there are 10 balls in total, the
probability that the first ball selected is blue is three out of 10 or
three-tenths. The probability that the first ball
is red is seven-tenths. In this question, we are told that
the first ball is red. This means that there are now six
red balls remaining in the bag. There are still three blue balls,
giving us a total of nine balls. The probability that the second
ball is red given that the first one is red is therefore equal to six-ninths. By dividing the numerator and
denominator of this fraction by three, this simplifies to two-thirds. The value of π₯ on the tree diagram
is therefore equal to two-thirds.

We can check this answer by
considering pairs of branches, as we know these must sum to one. As the probability that the second
ball is blue given that the first one is red is one-third and one-third plus
two-thirds equals one, we know that our answer is correct. The fraction one-third came from
the fact that three of the nine remaining balls are blue. And three-ninths is equivalent to
one-third.

Whilst it is not required in this
question, it is worth noting that when the first ball selected is blue, there will
be seven red and two blue balls left in the bag. This results in probabilities of
two-ninths and seven-ninths for the second ball being blue and red, respectively,
given that the first ball is blue.

In our next question, we will draw
a tree diagram in order to find the probability of a conditional event.

A bag contains 22 red balls and 15
black balls. Two balls are drawn at random. Find the probability that the
second ball is black given that the first ball is red. Give your answer to three decimal
places.

In this question, we are told that
two balls are drawn at random from a bag. One way of representing this is
using a tree diagram. We know that the first ball
selected can be either red or black. The same is true of the second
ball, giving us four possible combinations: red red, red black, black red, or black
black. As the two balls are drawn at the
same time, we can assume this is done without replacement. This means that we are dealing with
dependent events and conditional probability.

Conditional probability can be
written using the notation shown: the probability of π΅ given π΄. In this question, we were asked to
find the probability that the second ball is black given that the first ball is
red. This is the probability
corresponding to the branch highlighted in pink. As the events are dependent, we
will firstly need to calculate the probability that the first ball drawn is red.

There are 22 red balls and 15 black
balls in the bag. This means there are a total of 37
balls. And the probability that the first
ball selected is red is 22 out of 37. Whilst it is not required in this
question, we can also add to our tree diagram the probability that the first ball is
black. This is equal to 15 out of 37. Letβs now consider how many balls
are left in the bag if the first one drawn is red. There are now 21 red balls, and
there are still 15 black balls. This is a total of 36, and the
probability of selecting a black ball now is 15 out of 36. This is the probability we are
looking for. The probability that the second
ball is black given that the first ball is red is 15 out of 36.

Whilst we would often leave our
answer as a fraction, in this case, we are asked to give our answer to three decimal
places. 15 divided by 36 or the simplified
fraction five divided by 12 is equal to 0.4166 and so on. We can round this to three decimal
places, giving us an answer of 0.417. At this stage, it is worth
completing the remainder of the tree diagram. If the first ball selected is red,
the probability that the second ball is also red is 21 out of 36, as 21 of the
remaining balls are red. If we now assume that the first
ball drawn was black, there would be 22 red balls left and only 14 black balls. This means that the probability
that the second ball is red given that the first ball is black is 22 out of 36. And the probability that the second
ball is black given that the first ball is also black is 14 out of 36.

It is also worth checking at this
stage that the three pairs of fractions circled sum to one. We can do this before or after
simplifying the fractions.

We will now consider one final
example.

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.

We will now finish this video by
summarizing the key points. We saw in this video that when
there are a relatively small number of outcomes, a tree diagram is a useful way to
illustrate the probability of compound events. We saw that the sum of the
probabilities for each set of branches equals one. Likewise, the sum of the
probabilities of all the final outcomes equals one. When dealing with conditional
probability, we saw that the probability of π΄ intersection π΅ is equal to the
probability of π΅ given π΄ multiplied by the probability of π΄.