Conditional Probability

Here we will learn about conditional probability, including the definition of conditional probability and how to calculate probabilities when there are conditional elements.

There are also conditional probability worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.

What is conditional probability?

Conditional probability is the probability of an event occurring based on the occurrence of another event. Conditional probability is a fundamental aspect of probability theory.

Conditional probability questions often involve picking two objects from a set. This is  because once you have picked the first object, the probabilities change for the second pick, based on the outcome of the first pick.

For example,

If we have a bag containing 3 strawberry sweets and 4 lemon sweets.

The probabilities of picking each type of sweet are

P(Strawberry) = \frac{3}{7}

P(Lemon) =\frac{4}{7}

If we picked out one strawberry sweet, there would be 2 strawberry sweets and 4 lemon sweets left.

The probabilities would now be

P(Strawberry) = \frac{2}{6}

P(Lemon) =\frac{4}{6}

If instead we picked out one lemon sweet, there would be 3 strawberry sweets and 3 lemon sweets left.

The probabilities would now be

P(Strawberry) = \frac{3}{6}

P(Lemon) =\frac{3}{6}

The probabilities are calculated based on what has already occurred.

What is conditional probability?

What is conditional probability?

How to calculate conditional probability

In order to calculate conditional probability:

1. Identify the number of desired outcomes under the condition.
2. Identify the total number of outcomes under the condition.
3. Write the probability.
   
   \text{Probability }=\frac{\text{number of desired outcomes}}{\text{total number of outcomes}}

Explain how to calculate conditional probability

Conditional probability examples

Example 1: using a Venn diagram

The diagram below shows the results when 25 students were asked whether they can speak French or Spanish.

A student is chosen at random. Find the probability that the student can speak French given that they can speak Spanish.

1. Identify the number of desired outcomes under the condition.

The desired outcome here is that the student speaks French.

The condition is that they speak Spanish.

We need to identify the number of students who speak French and Spanish.

There are 10 students who speak French and Spanish.

This will be the numerator of our fraction.

2Identify the total number of outcomes under the condition.

The condition is that they speak Spanish so we need to consider all the students who speak Spanish.

There are 10+4=14 students who speak Spanish.

This will be the denominator of our fraction.

3Write the probability.

This probability can cancel to \frac{5}{7} .

Example 2: using a two-way table

The table below shows the hair and eye colour of a group of people.

A person from the group is chosen at random. Find the probability that the person has blonde hair given that they have blue eyes.

Example 3: probability of one event given another event has occurred

A bag contains 6 blue balls and 10 yellow balls. One ball is picked and it is yellow, it is not put back into the bag. A second ball is then picked. Find the probability that the second ball is yellow given that the first ball was yellow.

Example 4: probability of one event given another event

Lauren has a deck of cards. Lauren draws one card from the deck and does not replace it. The card she draws is an ace. Lauren then picks a second card. Find the probability that the second card is not an ace given the first card was an ace. 

Example 5: probability of two events using a tree diagram

There are 12 socks in a drawer. 4 are white and 8 are black. William picks one sock and then picks a second sock without replacing the first. Find the probability that both socks are white.

We can draw a tree diagram to represent the possible outcomes,

Since there are 12 socks to begin with, 4 of them white and 8 black, the probability of picking a white sock on the first pick is \frac{4}{12} and the probability of picking a black sock is \frac{8}{12}.

The probabilities for the second pick need to be calculated given the condition that the first pick has already occurred.

Let’s first assume that the first pick was a white sock.

Now let’s assume the first sock was black.

Write the probability.

\begin{aligned} &\text{Probability }=\frac{\text{number of desired outcomes}}{\text{total number of outcomes}}\\\\ &\text{P(White sock) }=\frac{4}{11}\\\\ &\text{P(Black sock) }=\frac{7}{11} \end{aligned}

We can fill the rest of the tree diagram in.

To find the joint probability that both socks are white we use 

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

\begin{aligned} \text{P(white and white)}&=\frac{4}{12} \times \frac{3}{11}\\\\ &=\frac{12}{132}\\\\ &=\frac{1}{11} \end{aligned}

Example 6: probability of two events using a tree diagram

Karam has a box of chocolates containing 14 milk chocolates and 6 white chocolates. Karam eats 2 chocolates. Find the probability that Karam eats two chocolates of the same type.

We can represent this using a tree diagram.

We need to work out the probabilities for the second chocolate.

If we assume the first chocolate was milk…

If we assume the first chocolate was white…

Write the probability.

\begin{aligned} &\text{Probability }=\frac{\text{number of desired outcomes}}{\text{total number of outcomes}}\\\\ &\text{P(milk) }=\frac{14}{19}\\\\ &\text{P(white) }=\frac{5}{19} \end{aligned}

We can fill the rest of the tree diagram in.

To find the probability that he eats two of the same chocolates, we need the probability that he eats two milk chocolates or two white chocolates. 

\begin{aligned} &\text{P(milk and milk)}=\frac{14}{20} \times \frac{13}{19} = \frac{182}{380}\\\\ &\text{P(white and white)}=\frac{6}{20} \times \frac{5}{19} = \frac{30}{380} \end{aligned}

The probability that Karam eats two of the same type of chocolate is

\frac{182}{380} + \frac{30}{380} = \frac{212}{380} = \frac{53}{95} .

Common misconceptions

• Not changing the probability for the second pick when picking two objects

For example, if you have a bag containing 3 blue balls and 7 yellow balls, the probability of picking a blue ball on the first pick is \frac{3}{10}.

If a blue ball has been picked on the first pick and not replced, then the probability of a blue ball on the second pick is \frac{2}{9} because there are 2 blue balls left and 9 balls altogether.

• Incorrect fraction work

Adding, subtracting, multiplying or dividing fractions incorrectly.

• Simplifying fractions

Usually when we have fractions in maths we simplify them. In probability questions it is usually better to leave fractions in their unsimplified form. This is because we often need to add fractions to solve probability problems.

Remember that fractions need a common denominator in order to add them. Check with the question if you are asked to give your final answer as a fraction in its simplified form or not.

The next lessons are

• Combined events probability
• Probability distribution
• Describing probability

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