### Video Transcript

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 𝐵 given 𝐴. 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. Let’s 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.