### Video Transcript

There are three red marbles and π
blue marbles in a bag. Two marbles are picked at
random. The probability to pick two red
marbles or two blue marbles is three-sevenths. Show that two π squared minus 11π
plus 12 is equal to zero.

This is a tree diagram question
with a twist. Usually weβre given the total
number of marbles or the probabilities that occur along each branch. However, in this case, weβre given
that the number of blue marbles is π. Letβs sketch the tree diagram out
and see what it looks like.

The two options when a marble is
chosen are red and blue. We can write R and B to show
this. The total number of marbles can be
found by adding the number of red marbles and the number of blue marbles. In this case, that gives us an
algebraic expression of three plus π.

There are three red marbles. So the probability of choosing a
red marble is three out of three plus π. Similarly, weβre given that there
are π blue marbles. So the probability of choosing a
blue marble the first time is π out of three plus π.

For the second marble, the choices
remain the same but the probabilities change. We know this because weβre not told
that the marbles are replaced, but that two are chosen at random. Letβs start then by looking at the
options after having chosen a red marble. Since weβve chosen one red marble
already, there must be two red marbles left in the bag. The total number of marbles has
also decreased by one. Three plus π minus one is equal to
two plus π. This means then that the
probability of choosing a second red marble is two out of two plus π.

Now, since weβve already chosen a
red marble, the number of blue marbles remains the same β thatβs π. But this time thereβs still one
less marble in the bag. So the total is two plus π. The probability of choosing a blue
marble here is π out of two plus π.

Now, letβs consider the options
after having chosen a blue marble. Since weβve chosen a blue marble,
that means the number of red marbles remains unchanged: itβs three out of two plus
π since once again the number of marbles is one less than at the start. The number of blue marbles now is
π minus one since we chose one blue marble the first time. This means that the probability of
choosing a second blue marble is given by π minus one over two plus π.

Now, we have a full tree diagram,
we can refer back to the question and see which outcomes weβre interested in. We want the probability of choosing
two red marbles or two blue marbles. Remember when we want to find the
probability of two events occurring, we multiply their probabilities together. That means the probability of
choosing a red and then another red is three over three plus π multiplied by two
over two plus π.

We can multiply the numerator of
the first fraction by the numerator of the second and the denominator of the first
fraction by the denominator of the second. That gives us six over three plus
π multiplied by two plus π. The probability of choosing a blue
marble and then a second blue marble is given by π over three plus π multiplied by
π minus one over two plus π.

Once again, we multiply the
numerator of the first fraction by the numerator of the second and the denominator
of the first fraction by the denominator of the second. The next rule weβre interested in
is the OR rule. If we want to find the probability
of one event or another event occurring, we add their probabilities.

In this case, weβll add the
probability of choosing two red marbles and the probability of choosing two blue
marbles. Thatβs six over three plus π
multiplied by two plus π plus π multiplied by π minus one over three plus π
multiplied by two plus π. Since the denominator of these two
fractions is the same, we can add the numerators, giving us six plus π multiplied
by π minus one all over three plus π multiplied by two plus π.

Now, weβre told that this
probability is equal to three-sevenths. Notice also that weβre trying to
create a quadratic equation equal to zero. That means we can make our
expression for the probability of getting two red marbles or two blue marbles equal
to three-sevenths. We can begin to rearrange this
equation by multiplying both sides by three plus π multiplied by two plus π and
next multiplying everything by seven. Our next step will be to expand all
the sets of brackets on both sides of the equation.

Weβve actually finished with our
tree diagram. So weβre gonna move all the working
out to the left side of our page to create a little bit more space for this. First, letβs expand three plus π
multiplied by two plus π using the FOIL method. Three multiplied by two is six,
three multiplied by π is three π, π multiplied by two is two π, and π
multiplied by π is π squared. Collecting like terms gives us six
plus five π plus π squared. The entire expression then needs to
be multiplied by three, giving us 18 plus 15π plus three π squared.

Next, letβs expand the bracket π
multiplied by π minus one. π multiplied by π is π squared
and π multiplied by negative one is negative π. We can replace π multiplied by π
minus one then with π squared minus π. Then, we need to multiply
everything in this bracket by seven. That gives us 42 plus seven π
squared minus seven π.

Our last step is to rearrange so
that our equation is equal to zero. We start by subtracting three π
squared from both sides, then we subtract 15π from both sides, and finally we
subtract that 18 from both sides. That leaves us with 24 plus four π
squared minus 22π is equal to zero. Each term in this equation can be
divided by two. Dividing through by two and
rearranging the expression so that the powers of π are descending gives us two π
squared minus 11π plus 12 is equal to zero.

Find the probability to pick two
marbles of different colour.

If we recall the tree diagram for
part a of the question, there were four possible outcomes: we could get a red and a
red marble, we could get a red and a blue, a blue then a red, or finally we could
get two blue marbles. To pick two marbles of different
colour, weβre interested in red and blue or blue and red.

Now remember the probabilities of
all possible outcomes from an experiment must sum to one. We already know the probability of
getting two red marbles or two blue marbles is three-sevenths. We can therefore subtract
three-sevenths from one whole to give us the probability of getting red and blue or
blue and red. One minus three-sevenths is
four-sevenths.

So the probability to pick two
marbles of different colour is four-sevenths.