Video Transcript
A bag contains 27 white balls and
six black balls. If two balls are drawn
consecutively without replacement, what is the probability that the second ball is
black given that the first is black?
As this question concerns two balls
being drawn one after the other from a bag, it will be helpful to use a tree diagram
to illustrate the situation. The first event is selecting the
first ball which has two outcomes: it is either white or black. The second event, which is
selecting the second ball, has the same two outcomes: it is also either white or
black.
Notice that the two balls are drawn
without replacement, so once the first ball has been selected, it is not put back in
the bag before the second ball is drawn. We therefore need to consider the
outcome of the color of the first ball when calculating the probability of each
outcome for the second ball. For the first ball though, we can
just use the original number of balls of each color in the bag. Initially, there are 27 white balls
and six black balls. So the total number of balls in the
bag is 33. We can therefore write the
probabilities for the outcome of the first ball as fractions with a denominator of
33.
As there are 27 white balls
initially, the probability that the first ball chosen is white is 27 out of 33. And as there are six black balls
initially, the probability that the first ball chosen is black is six out of 33. Now we need to consider the
probabilities for the outcome of the second ball selection. As the first ball has not been
replaced, there is now one less ball in the bag. So the probabilities for the second
ball can all be written as fractions with a denominator of 32.
Now we need to consider carefully
what has already happened when determining the numerator for each fraction. If the first ball taken was white,
then there is now one less white ball in the bag, so there are 26 white balls
remaining. The probability that the second
ball is white, if the first was white, is therefore 26 out of 32. If the first ball was white, then
the number of black balls is unchanged. So there are still six black balls
in the bag. The probability the second ball is
black, if the first ball was white, is therefore six out of 32.
If on the other hand the first ball
chosen was black, then there is now the same number of white balls in the bag as
when we started, but the number of black balls would have reduced by one. So the probability the second ball
is white, if the first was black, is 27 out of 32. And the probability the second ball
is black, if the first was black, is five out of 32.
Notice that in each case the sum of
the probabilities on each set of branches of the tree diagram is one, which can be
helpful in determining the probabilities.
Now let’s use our tree diagram to
answer the question, which was to find the probability that the second ball is black
given that the first is black. This probability can be found on
the second set of branches of the tree diagram. Given that the first ball is black,
the probability that the second ball will also be black is five out of 32, because
if one black ball has been removed from the bag, there are five black balls
remaining out of the reduced total of 32. We can express this using the
conditional probability notation of a vertical line.
So, by drawing a tree diagram,
we’ve found that the probability the second ball is black, given that the first is
black, is five over 32.
