8.04 Conditional probability

We are often interested in probabilities under certain conditions. For example what is the probability of it snowing tomorrow given the temperature today, or if an even number is rolled on a die what is the probability that it is a six. Giving conditions will reduce the sample space we are concerned with.

For two events A and B the probability of event A occurring given that event B has occurred has the notation P(A|B) which is read as the "probability of A given B". This given by: P(A|B)=\dfrac{P(A \cap B)}{P(B)}.

If we know the conditional probability we can calculate the probability of A and B using either: \\ P(A \cap B)=P(A|B)P(B) or P(A \cap B) = P(B|A)P(A).

Examples

Recall that independent events are events where the occurrence of one event does not affect the probability of the other occurring. Such as rolling a dice and then tossing a coin, or tossing a coin repeatedly. Now in the context of conditional probability this would mean that the probability of event A occurring given event B has occurred should simply be the probability of A - that is, it shouldn't matter if event B occurred or not. Mathematically we can write this property as: P(A|B)=P(A).

From our conditional formula we see that for independent events:

We used this fact last lesson to calculate the probability for P(A \cap B). We can now also use this fact to test if two events are independent if we know the probability of A,\,B and the probability of (A and B).

Examples

Example 2

The probability of two independent events, A and B are, P(A)=0.6 and P(B)=0.7.

Determine the probability of:

a

Both A and B occurring.

Worked Solution

Create a strategy

We can use the formula: P(A \cap B) = P(A) \times P(B).

Apply the idea

\displaystyle P(A \cap B)	\displaystyle =	\displaystyle 0.6 \times 0.7	Substitute the probabilities
	\displaystyle =	\displaystyle 0.42	Evaluate

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b

Neither A nor B.

Worked Solution

Create a strategy

We can use the formula: P(A' \cap B') = P(A') \times P(B').

Apply the idea

\displaystyle P(A' \cap B')	\displaystyle =	\displaystyle (1-P(A))(1-P(B))	Find the complement probabilities
\displaystyle P(A' \cap B')	\displaystyle =	\displaystyle (1-0.6)(1-0.7)	Substitute P(A) = 0.6 and P(B)=0.7
	\displaystyle =	\displaystyle 0.4 \times 0.3	Evaluate the subtraction
	\displaystyle =	\displaystyle 0.12	Evaluate the multiplication

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c

A or B or both.

Worked Solution

Create a strategy

This is the complement of neither A nor B.So we can use the formula: P(A \cup B)= 1-P(A' \cap B').

Apply the idea

We get from part (b) that P(A' \cap B') = 0.12.

\displaystyle P(A \cup B)	\displaystyle =	\displaystyle 1-0.12	Substitute P(A' \cap B')=0.12
	\displaystyle =	\displaystyle 0.88	Evaluate the subtraction

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d

B but not A.

Worked Solution

Create a strategy

We can use the formula: P(B \cap A') = P(B) \times P(A').

Apply the idea

\displaystyle P(B \cap A')	\displaystyle =	\displaystyle 0.7(1-0.6)	Substitute P(A)=0.6,\, P(B)=0.7
	\displaystyle =	\displaystyle 0.7 \times 0.4	Evaluate the subtraction
	\displaystyle =	\displaystyle 0.28	Evaluate the multiplication

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e

A given that B occurs.

Worked Solution

Create a strategy

Since the two events are independent, we can use the formula : P(A|B) = P(A).

Apply the idea

\displaystyle P(A|B)

\displaystyle =

\displaystyle 0.6

Substitute P(A)

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Example 3

A flight departs from Melbourne to Sydney. The probability that the flight departs on time, given the weather is fine in Melbourne is 0.9, and the probability that the flight departs on time, given the weather is not fine in Melbourne is 0.7. The probability that the weather is fine on any particular day in July is 0.4.

a

By constructing a tree diagram or otherwise, find the probability that the flight from Melbourne to Sydney departs on time in a day in July.

Worked Solution

Create a strategy

Use a tree diagram.

Apply the idea

We can construct the following tree diagram for this situation:

The flight can be on time in two different ways depending on whether the weather is fine: P(\text{On time})=P(\text{Fine}) \times P(\text{On time}|\text{Fine}) + P(\text{Not fine}) \times P(\text{On time}|\text{Not fine})

\displaystyle P(\text{On time})	\displaystyle =	\displaystyle 0.4 \times 0.9 + 0.6 \times 0.7	Substitute the probabilities
	\displaystyle =	\displaystyle 0.78	Evaluate

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b

Find the probability that the weather is fine in Melbourne given that the flight departs on time on a day in July?

Worked Solution

Create a strategy

We can use the formula: P(A|B)=\dfrac{P(A \cap B)}{P(B)}.

Apply the idea

\displaystyle P(\text{Fine | On time})	\displaystyle =	\displaystyle \dfrac{P(\text{Fine}\cap \text{On time})}{P(\text{On time})}	Write the formula
\displaystyle P(\text{Fine | On time})	\displaystyle =	\displaystyle \dfrac{0.4\times 0.9}{0.78}	Substitute the probabilities
	\displaystyle =	\displaystyle \dfrac{6}{13}	Evaluate

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