5.3.3 Tree Diagrams and Conditional Probability

In this topic we will learn how to:

• calculate and use conditional probabilities in simple cases

Tree diagrams are a way to calculate the possible outcomes in an organized way. We will learn how to draw and use them below.

Conditional probability is the likelihood of an event occurring based on the occurrence of a previous event. It is denoted by,
P(A | B)The above reads as the probability of A given B where A and B are two different events. The formula for calculating conditional probability is,
P(B | A) = \frac{P(A \cap B)}{P(A)}Let’s look at some past paper questions.

1. On Mondays, Rani cooks her evening meal. She has a pizza, a burger or a curry with probabilities 0.35, 0.44, 0.21 respectively. When she cooks a pizza, Rani has some fruit with a probability of 0.3. When she cooks a burger, she has some fruit with a probability of 0.8. When she cooks a curry, she never has any fruit. (9709/52/M/J/21 number 3)

(a) Draw a fully labeled tree diagram to represent this information.

Start by drawing three branches,

Label each branch with Pizza, Burger, Curry, and the respective probabilities,

Draw branches extending from first set of branches labelled with Fruit or No Fruit, for each of Pizza, Burger and Curry and label with respective probabilities. This will complete the tree diagram, therefore, the final answer is,

(b) Find the probability that Rani has some fruit.

The probability that Rani has some fruit, can be written as,
P(\textmd{Fruit}) = P(\textmd{Pizza and Fruit}) + P(\textmd{Burger and Fruit}) + P(\textmd{Curry and Fruit})Read off the probabilities from the tree diagram and substitute,
P(\textmd{Fruit}) = (0.35 \times 0.3) + (0.44 \times 0.8) + (0.21 \times 0)This simplifies to give,
P(\textmd{Fruit}) = 0.105 + 0.352P(\textmd{Fruit}) = 0.457Therefore, the final answer is,
P(\textmd{Fruit}) = 0.457(c) Find the probability that Rani does not have a burger given that she does not have any fruit.

In the question they have used the word ‘given’, this implies that this is conditional probability. The formula for conditional probability is,
P(B | A) = \frac{P(A \cap B)}{P(A)}Let’s assign the letters to their events in context,
\textmd{A - Rani does not have fruit}\textmd{B - Rani does not have a burger}\textmd{A} \cap\textmd{ B - Rani does not have fruit and does not have a burger}Note: In the formula, it is B given A. So from the question, assign the ‘given’ event to A, and the other event to B.

Now let’s find the probability of event A happening. In part (b) we calculated the probability that Rani has fruit, we can use that to find the probability that he does not have fruit,
P(A) = 1 - P(\textmd{Fruit})P(A) = 1 - 0.457P(A) = \textcolor{#2192ff}{0.543}Now let’s find the probability of the intersection of A and B. This can be written as,
P(A \cap B) = P(\textmd{No Burger and No Fruit})P(A \cap B) = P(\textmd{Pizza and No Fruit}) + P(\textmd{Curry and No Fruit})Read off the probabilities from the tree diagram,
P(A \cap B) = (0.35 \times 0.7) + (0.21 \times 1)This simplifies to give,
P(A \cap B) = 0.245 + 0.21P(A \cap B) = \textcolor{#0f0}{0.455}Now let’s substitute into the formula for conditional probability,
P(B | A) = \frac{P(A \cap B)}{P(A)}P(B | A) = \frac{\textcolor{#0f0}{0.455}}{\textcolor{#2192ff}{0.543}}P(B | A) = \frac{455}{543}P(B | A) = 0.838Therefore, the final answer is,
P(B | A) = 0.8382. On each day that Alexa goes to work, the probabilities that she travels by bus, by train or by car are 0.4, 0.35 and 0.25 respectively. When she travels by bus, the probability that she arrives late is 0.55. When she travels by train, the probability that she arrives late is 0.7. When she travels by car, the probability that she arrives late is x. On a randomly chosen day when Alexa goes to work, the probability that she does not arrive late is 0.48. (9709/52/M/J/21 number 3)

(a) Find the value of x.

Let’s sketch a tree diagram, with the information we have been given,

We have been given that,
P(\textmd{Not Late}) = \textcolor{#2192ff}{0.48}We will use that idea to solve for x.

Using the tree diagram, we can also say that,

P(\textmd{Not Late}) = P(\textmd{Bus and Not Late}) + P(\textmd{Train and Not Late}) + P(\textmd{Car and Not Late})

Therefore, we can rewrite that as,
\textcolor{#2192ff}{0.48} = P(\textmd{Bus and Not Late}) + P(\textmd{Train and Not Late}) + P(\textmd{Car and Not Late})Read off the probabilities from the tree diagram,
0.48 = (0.4 \times 0.45) + ( 0.35 \times 0.3) + [0.25(1 - x)]Simplify and solve for x,
0.48 = 0.18 + 0.105 + 0.25 - 0.25x0.48 = 0.535 - 0.25x0.25x = 0.535 - 0.480.25x = 0.055x = 0.22Therefore, the final answer is,
x = 0.22(b) Find the probability that Alexa travels to work by train given that she arrives late.

In the question they have used the word ‘given’, this implies that this is conditional probability. The formula for conditional probability is,
P(B | A) = \frac{P(A \cap B)}{P(A)}Let’s assign the letters to their events in context,
\textmd{A - Alexa arrives late}\textmd{B - Alexa travels to work by train}\textmd{A} \cap\textmd{ B - Alexa travels by train and arrives late}Note: In the formula, it is B given A. So from the question, assign the ‘given’ event to A, and the other event to B.

Now let’s find the probability of event A happening. In the stem of the question we’re given that the probability that Alexa does not arrive late is 0.48, let’s use that to calculate the probability that she does arrive late,
P(A) = 1 - P(\textmd{Not Late})P(A) = 1 - 0.48P(A) = \textcolor{#2192ff}{0.52}Now let’s find the probability of the intersection of A and B. This can be written as,
P(A \cap B) = P(\textmd{Train and Late})Read off the probabilities from the tree diagram,
P(A \cap B) = 0.35 \times 0.7This simplifies to give,
P(A \cap B) = \textcolor{#0f0}{0.245}Now let’s substitute into the formula for conditional probability,
P(B | A) = \frac{P(A \cap B)}{P(A)}P(B | A) = \frac{\textcolor{#0f0}{0.245}}{\textcolor{#2192ff}{0.52}}P(B | A) = \frac{49}{104}Therefore, the final answer is,
P(B | A) = \frac{49}{104}