Video: Pack 3 β€’ Paper 2 β€’ Question 22

Pack 3 β€’ Paper 2 β€’ Question 22

07:47

Video Transcript

There are three red marbles and 𝑏 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 𝑏 squared minus 11𝑏 plus 12 is equal to zero.

This is a tree diagram question with a twist. Usually we’re given the total number of marbles or the probabilities that occur along each branch. However, in this case, we’re given that the number of blue marbles is 𝑏. Let’s 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 𝑏.

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

For the second marble, the choices remain the same but the probabilities change. We know this because we’re not told that the marbles are replaced, but that two are chosen at random. Let’s start then by looking at the options after having chosen a red marble. Since we’ve 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 𝑏 minus one is equal to two plus 𝑏. This means then that the probability of choosing a second red marble is two out of two plus 𝑏.

Now, since we’ve already chosen a red marble, the number of blue marbles remains the same β€” that’s 𝑏. But this time there’s still one less marble in the bag. So the total is two plus 𝑏. The probability of choosing a blue marble here is 𝑏 out of two plus 𝑏.

Now, let’s consider the options after having chosen a blue marble. Since we’ve chosen a blue marble, that means the number of red marbles remains unchanged: it’s three out of two plus 𝑏 since once again the number of marbles is one less than at the start. The number of blue marbles now is 𝑏 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 𝑏 minus one over two plus 𝑏.

Now, we have a full tree diagram, we can refer back to the question and see which outcomes we’re 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 𝑏 multiplied by two over two plus 𝑏.

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 𝑏 multiplied by two plus 𝑏. The probability of choosing a blue marble and then a second blue marble is given by 𝑏 over three plus 𝑏 multiplied by 𝑏 minus one over two plus 𝑏.

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 we’re 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, we’ll add the probability of choosing two red marbles and the probability of choosing two blue marbles. That’s six over three plus 𝑏 multiplied by two plus 𝑏 plus 𝑏 multiplied by 𝑏 minus one over three plus 𝑏 multiplied by two plus 𝑏. Since the denominator of these two fractions is the same, we can add the numerators, giving us six plus 𝑏 multiplied by 𝑏 minus one all over three plus 𝑏 multiplied by two plus 𝑏.

Now, we’re told that this probability is equal to three-sevenths. Notice also that we’re 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 𝑏 multiplied by two plus 𝑏 and next multiplying everything by seven. Our next step will be to expand all the sets of brackets on both sides of the equation.

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

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

Our last step is to rearrange so that our equation is equal to zero. We start by subtracting three 𝑏 squared from both sides, then we subtract 15𝑏 from both sides, and finally we subtract that 18 from both sides. That leaves us with 24 plus four 𝑏 squared minus 22𝑏 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 𝑏 are descending gives us two 𝑏 squared minus 11𝑏 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, we’re 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.

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