 Probability and Tree Diagrams Worksheets | Questions and Revision

# Probability and Tree Diagrams Worksheets, Questions and Revision

Level 4 Level 5

## Probability & Tree Diagrams

Probability is the study of how likely things are to happen. This topic will look at how tree diagrams, can be used to determine the probability of different types of events happening.

Make sure you are happy with the following topics before continuing.

## Independent and Dependent Events

We need to understand independent and dependent events to be able to do the next sections.

• Two or more events are independent if one event doesn’t effect the probability of the others happening.
• Two or more events are dependent if one event does effect the probability of the others happening.

Example:

• Getting a head both times on $2$ coin flips are independent events.
• Picking a red marble then picking a green marble without replacing the red marble are dependent events.
Level 6-7

## The AND Rule

• The AND rule states that: If two events, $A$ and $B$, are independent, then,

$\textcolor{black}{\text{P}(A \text{ and } B) = \text{P}(A) \times \text{P}(B)}$

Level 6-7

## The OR Rule

• The OR rule states that: for two events, $A$ and $B$, then,

$\textcolor{black}{\text{P}(A \text{ or } B) = \text{P}(A) + \text{P}(B) - \text{P}(A \text{ and } B)}$

• If $A$ and $B$ cannot happen together, we say they are mutually exclusive, and then we have $\text{P}(A \text{ and } B) = 0$, so the OR rule becomes

$\textcolor{black}{\text{P}(A \text{ or } B) = \text{P}(A) + \text{P}(B)}$

Level 6-7

## Probability Trees Probability trees are similar to frequency trees, but we instead put the probabilities on the branches and the events at the end of the branch.

Example: A bag contains $4$ red balls and $5$ blue balls. Raheem picks $2$ balls at random. Calculate the probability that he selects the same coloured ball each time, given that after each time a ball is selected, it is replaced

• Step 1: Construct the probability tree showing two selections, noting that there are $9$ balls in total and a $\dfrac{4}{9}$ chance of picking a red ball. Then as the red ball is replaced, there are still $4$ red balls left out of $9$, so again there is a $\dfrac{4}{9}$ chance of picking a red ball on the second selection.
• Step 2: Use the tree diagram to determine the probability of selecting the same colour twice. We can see that there are two ways of doing this either blue and blue, or, red and red. We use the AND rule via the tree diagram, so

$\textcolor{black}{\text{P(blue and blue)}=\dfrac{5}{9}\times\dfrac{5}{9}= \textcolor{blue}{\dfrac{25}{81}} \quad \text{ and } \quad \text{P(red and red)}=\dfrac{4}{9}\times\dfrac{4}{9}= \textcolor{red}{\dfrac{16}{81}} }$

• Step 3: The final step then is to add the probabilities together, by the OR rule for mutually exclusive events, to get,

$\textcolor{black}{\text{P(same colour)}= \dfrac{25}{81} +\dfrac{16}{81}=\dfrac{41}{81}}$

Level 6-7

## Conditional Probability

The conditional probability of $A$ given $B$, is the “probability that event $A$ happens given that event $B$ happens”. You will not be told that it is a conditional probability question, but seeing words like ‘without replacement’ or ‘given’ will mean that it is one, or you may have to use your own intuition.

• If two events, $A$ and $B$, are independent, then

$\textcolor{black}{\text{P}(A \text{ given } B) = \text{P}(B) \quad \text{ and } \quad \text{P}(B \text{ given } A) = \text{P}(A)}$

• If two events, $A$ and $B$ are dependent, then

$\textcolor{black}{\text{P}(A \text{ and } B) = \text{P}(A) \times \text{P}(B \text{ given } A)}$

Example:

Benjamin plays football for his local team. The probability that he is in the starting line up for his team this Sunday is $0.7$. If he starts the game, the probability that he scores a goal is $0.4$. What is the probability that Benjamin starts the game but doesn’t score a goal?

• Step 1: We want to find $\text{P(starts and doesn't score)}$

Let “starts” be event $A$ and “doesn’t score” be event $B$

• Step 2: $\text{P}(A) = 0.7$

$\text{P}(B \text{ given } A) = \text{P(doesn't score given he starts)} = 1 - 0.4 = 0.6$

• Step 3: Then,

$\text{P}(A \text{ and } B) = \text{P}(A) \times \text{P}(B \text{ given } A) = 0.7 \times 0.6 = 0.42$

Level 8-9

## Conditional Probability Trees Conditional probability trees are similar to probability trees, but the probabilities change depending on the previous events.

Example: A bag contains $4$ red balls and $5$ blue balls. Raheem picks $2$ balls at random. Calculate the probability that he selects the same coloured ball each time, given that each time a ball is selected, it is not replaced

• Step 1: Construct the probability tree showing two selections, noting that there are $9$ balls to begin with, reducing to $8$ after the first selection, as shown below, e.g. there is a $\dfrac{4}{9}$ chance of picking a red ball, then since a red ball has been removed and not replaced, there are only $3$ red balls left out of $8$, so there is a $\dfrac{3}{8}$ chance of picking a red ball on the second selection, and so on.
• Step 2: Use the tree diagram to determine the probability of selecting the same colour twice. We can see that there are two ways of doing this either blue and blue, or, red and red. We use the AND rule via the tree diagram, so

$\textcolor{black}{ \text{P(blue and blue)}=\dfrac{5}{9}\times\dfrac{4}{8}= \textcolor{blue}{\dfrac{20}{73}} \text{and} \text{P(red and red)}=\dfrac{4}{9}\times\dfrac{3}{8}= \textcolor{red}{\dfrac{12}{73}} }$

• Step 3: The final step then is to add the probabilities together, by the OR rule for mutually exclusive events, to get

$\textcolor{black}{\text{P(Same colour)}= \dfrac{20}{73} +\dfrac{12}{73}=\dfrac{32}{73}}$

Level 8-9

### Example Questions

(a) Let “Anna passing” be event $A_p$ and “Rob” passing be event $B_p$.

To work out the probability of Rob passing we can write the probability of both passing as:

$P(A_p \text{ and } R_p) =0.35$

Substituting in the probability of Anna passing her test,

$0.7 \times P(R_p) =0.35$

Rearranging the equation to make $P(R_p)$ the subject:

$P(R_p) = 0.35 \div 0.7 = 0.5$

(b)  The probability of both Anna and Rob failing their driving test can be found using a tree diagram as shown below: Hence the probability of them both failing is $\dfrac{3}{20} = 0.15$.

For this question when drawing the tree diagram we have to be careful as the probability changes between the two events. This is the result of not replacing the first counter hence only leaving $11$ counters in the bag to pick from. Adding together the probabilities of the result being blue then blue or green then green:

$\dfrac{7}{22}+\dfrac{5}{33}=\dfrac{31}{66}$

To work out the probability of the bus being late on both of the days we can use a tree diagram where $E$ represents the bus being on time or early and $L$ represents the bus being late. Going bottom line we find that the probability of being late of both days is:

$\dfrac{1}{16}$

Here we have to work out the probability that the coach takes out two balls that are a different colour.

For conditional probability questions, when drawing the tree diagram we have to be careful as the probability changes between the two events. This is the result of not replacing the first ball hence only leaving $13$ balls in the bag to pick from. Adding together the probabilities of the result being two different colours:

$\dfrac{45}{182}+\dfrac{45}{182}=\dfrac{90}{182}=\dfrac{45}{91}$

Hence as this is just below a half it is more likely that the coach picks two balls that are of the dame colour.

(a) The resultant tree diagram should look something like: (b) To find the probability he wins at least one game we can simple add the top $3$ branches probabilities together or subtract the probability of bottom branch from $1$

$\dfrac{9}{25}+\dfrac{6}{25}+\dfrac{6}{25}=\dfrac{21}{25}$

or,

$1-\dfrac{4}{25}=\dfrac{21}{25}$

Level 4-5

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