Apply Tree Diagrams with Algebra Worksheet

So far, we have looked at using tree diagrams to find the outcomes of multiple events when we are given the probability of a single event.

Now we need to look at doing this the other way round.

Example 1: A bag contains red and blue balls. A ball is picked out, the colour noted and then replaced. This is repeated.

If the probability of picking out two red balls is 1/25. What is the probability of picking out two blue balls.

Step 1: Lets put this into a tree diagram.

Step 2: What can we work out?

We know that the probabilities are the same for each event because we are told the ball is replaces.

If we know that P_(AA) = 1/25, this is the same as saying P_(A) x P_(A)= 1.25 so P_(A) = 1/5

Step 3: Work out the other event.

We know that A and B are exhaustive, so P_(B) must equal 4/5

Step 4: Complete the rest of the tree diagram.

Step 5: Use this to answer the question.

We were initially asked for the Probability of 2 blue balls. From the tree diagram, we can now see that this is 16/25

Complete the sentence below.

For this question, draw a tree diagram before you answer the question.

A bag contains only blue and red balls.

A ball is drawn and the colour noted. It is then replaced and another is taken out.

If the probability of drawing two blue balls is 9/49, find the probability of drawing two reds.

For this question, draw a tree diagram before you answer the question.

A bag contains only blue and red balls.

A ball is withdrawn and the colour noted. It is then replaced and another is taken out.

If the probability of getting 2 blue balls is 9/49, find the probability of getting ONLY 1 red ball during two draws.

12/49

24/49

18/49

For this question, draw a tree diagram before you answer the question.

A bag contains only blue and red balls.

A ball is withdrawn and the colour noted. It is replaced and another is taken out.

If the probability of getting 2 blue balls is 9/49, find the probability of getting AT LEAST 1 red ball during two draws.

12/49

24/49

18/49

A bag contains 10 sweets. Some are mint and some are orange.

Lionel takes two sweets.

For the tree diagram shown below, which of the options shows the correct algebraic terms for A and B?

A =

x

10

and B =

x

9

A =

x

10

and B =

x-1

9

A bag contains 10 sweets.

Some are mint and some are orange.

Lionel takes two sweets.

The diagram below expresses these outcomes visually:

In the previous question, we worked out the two individual probabilities for mint outcomes:

P(M) = x/10
P(MM) = (x - 1)/9

Use this information to find the total number of mint sweets in the bag at the start.

A =

x

10

and B =

x

9

A =

x

10

and B =

x-1

9

A bag contains 10 sweets.

Some are mint and some are orange.

Lionel takes two sweets.

We have now found that there were 8 mint sweets in the bag originally (out of 10).

What is the probability that Lionel doesn't get a mint sweet on either draw?

A =

x

10

and B =

x

9

A =

x

10

and B =

x-1

9

A bag contains red and blue balls.

If P(R) = a, what is P(B)?

Don't use any spaces between characters in your answer.

A =

x

10

and B =

x

9

A =

x

10

and B =

x-1

9

A bag contains red and blue balls.

P(R) = a

P(B) = 1 - a

Imagine we take out two balls in a row from the bag.

Match each probability on the left with their algebraic forms on the right.

Column A	Column B
P_(RR)	a²
P_(RB)	a(1 - a)
P_(BR)	a(1 - a)
P_(BB)	1 - a²

A bag contains red and blue balls.

P(R) = a

P(B) = 1 - a

Imagine we take out two balls in a row from the bag, replacing them each time.

Find the probability that exactly one of the balls will be blue.

Give your answer in its simple factorised form.

Don't use any spaces between characters in your answer.

ANSWERS

Question 1

Complete the sentence below.

CORRECT ANSWER

EDDIE SAYS

You approach tree diagrams by working backwards. Once you are given the outcomes, you can work backwards using algebra to then calculate the rest of the table.

ANSWERS

Question 2

For this question, draw a tree diagram before you answer the question.

A bag contains only blue and red balls.

A ball is drawn and the colour noted. It is then replaced and another is taken out.

If the probability of drawing two blue balls is 9/49, find the probability of drawing two reds.

CORRECT ANSWER

EDDIE SAYS

We know that P(BB) = 9/49 If we work backwards and find the square of both numbers, we can work out that P(B) = 3/7 For the first draw, we know that the overall of the two outcomes (red or blue) must equal 1. So we can know that P(R) = 4/7 (1 - 3/7) All we have to do now is find the probability of getting a red and then a red: P(RR) = 4/7 × 4/7 = 16/49

ANSWERS

Question 3

For this question, draw a tree diagram before you answer the question.

A bag contains only blue and red balls.

A ball is withdrawn and the colour noted. It is then replaced and another is taken out.

If the probability of getting 2 blue balls is 9/49, find the probability of getting ONLY 1 red ball during two draws.

CORRECT ANSWER

24/49

EDDIE SAYS

We know that P(BB) = 9/49 This means that P(B) = 3/7 From this, we can find that P(R) = 4/7 There are two ways we can obtain only one red from two draws: RB and BR. We need to find these two probabilities and then add them together: P(RB) = 4/7 × 3/7 = 12/49 P (BR) = 3/7 × 4/7 = 12/49 P(RB or BR) = 12/49 + 12/49 = 24/49

ANSWERS

Question 4

For this question, draw a tree diagram before you answer the question.

A bag contains only blue and red balls.

A ball is withdrawn and the colour noted. It is replaced and another is taken out.

If the probability of getting 2 blue balls is 9/49, find the probability of getting AT LEAST 1 red ball during two draws.

CORRECT ANSWER

EDDIE SAYS

We know that P(BB) = 9/49 This means that P(B) = 3/7 From this, we can find that P(R) = 4/7 There are three ways to obtain at least one red from two draws: RB, BR and RR. We need to find these three probabilities and then add them together: P(RB) = 4/7 × 3/7 = 12/49 P(BR) = 3/7 × 4/7 = 12/49 P(RR) = 4/7 × 4/7 = 16/49 P(RB, BR and RR) = 12/49 + 12/49 + 16/49 = 40/49 You could also use the fact that P(at least) = 1 - P(none) P(at least 1 red) = 1 - P(BB) This will give the same answer and be a bit quicker.

ANSWERS

Question 5

A bag contains 10 sweets. Some are mint and some are orange.

Lionel takes two sweets.

For the tree diagram shown below, which of the options shows the correct algebraic terms for A and B?

CORRECT ANSWER

A =

x

10

and B =

x-1

9

EDDIE SAYS

This is a conditional probability question, so we have to reduce the number of possible outcomes in the second scenario by 1 to show that the sweet has not been replaced. As we don't know how many mints there are, we will call this value 'x'. The probability of getting a mint on the second draw too will be 1 less than the original, so we will call this 'x - 1'. The question tells us that there are 10 starting outcomes, which will reduce to 9 for the second draw. Can you put these facts together to select the correct answer?

ANSWERS

Question 6

A bag contains 10 sweets.

Some are mint and some are orange.

Lionel takes two sweets.

The diagram below expresses these outcomes visually:

In the previous question, we worked out the two individual probabilities for mint outcomes:

P(M) = x/10
P(MM) = (x - 1)/9

Use this information to find the total number of mint sweets in the bag at the start.

CORRECT ANSWER

EDDIE SAYS

We can see that P(MM) = 28/45 We can use this information to create a quadratic formula: x/10 × (x - 1)/9 = 28/45 Rearranging this gives us: (x² - x)/90 = 28/45 Cross multiply both sides to reach: 45(x² - x) = 2520 Divide both sides by 45: x² - x = 56 x² - x - 56 = 0 We can solve this to find that x = 8 or x = -7. Which one of these has to be correct in this case?

ANSWERS

Question 7

A bag contains 10 sweets.

Some are mint and some are orange.

Lionel takes two sweets.

We have now found that there were 8 mint sweets in the bag originally (out of 10).

What is the probability that Lionel doesn't get a mint sweet on either draw?

CORRECT ANSWER

EDDIE SAYS

For this, we're now looking for P(OO). The probability of the first sweet Lionel picks being orange is 2/10. The probability of the second being orange, after the first one was too, is 1/9. Now we need to multiply these together to find the probability of both these outcomes: 2/10 × 1/9 = 2/90 Which simplifies to: 1/45

ANSWERS

Question 8

A bag contains red and blue balls.

If P(R) = a, what is P(B)?

Don't use any spaces between characters in your answer.

CORRECT ANSWER

EDDIE SAYS

The most important fact here is that the bag only contains red and blue balls. This means the probabilities of red and blue together must add to 1. P(R) + P(B) = 1 P (B) = 1 - P(R) P(B) = 1 - a

ANSWERS

Question 9

A bag contains red and blue balls.

P(R) = a

P(B) = 1 - a

Imagine we take out two balls in a row from the bag.

Match each probability on the left with their algebraic forms on the right.

CORRECT ANSWER

Column A	Column B
P_(RR)	a²
P_(RB)	a(1 - a)
P_(BR)	a(1 - a)
P_(BB)	1 - a²

EDDIE SAYS

If we apply the probabilities we found in the previous question, we need to multiply these together in the correct way for the outcome expressed. e.g. P(RR) means the probability of getting a red ball then another red ball. P(RR) = P(R) × P(R) P(RR) = a × a P(RR) = a² Can you apply this logic to find the remaining matches independently?

ANSWERS

Question 10

A bag contains red and blue balls.

P(R) = a

P(B) = 1 - a

Imagine we take out two balls in a row from the bag, replacing them each time.

Find the probability that exactly one of the balls will be blue.

Give your answer in its simple factorised form.

Don't use any spaces between characters in your answer.

CORRECT ANSWER

2a(1-a)

EDDIE SAYS

There are two ways in which we can get exactly one blue ball: RB which has a probability of a(1 - a) = a - a² BR which has a probability of (1 - a)a = a - a² Adding these together gives us: (a - a²) + (a - a²) 2a - 2a² If we factorise (add brackets) to this, we reach: 2a(1 - a) A tricky one to finish on there - well done!