Conditional probability
This article explores solving problems using conditional probability.
An event B may depend on event A that has already occurred. Consider the following Venn diagram.
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The event A has already happened, so we’re considering probabilities inside the closed curve A. We are looking for the probability of B given A has happened so we divide i by P(A) to rescale the probability of event A to 1. So the probability of;
The probability of B given A written P(B|A) is called the conditional probability of B given;
We can derive a formula known as the multiplication rule;
Note that we can swap A and B around to get;
Examples
Two fair spinners each have four faces numbered 1 to 4, The two spinners are thrown together and the sum of the numbers indicated on each spinner is recorded.
Given that at least one spinner lands on a 3, find the probability of the spinners indicating a sum of exactly 5.
We will need to draw two axes to indicate the values on each spinner. Let A stand for the event ‘at least one spinner lands on a 3’ and B the event ‘the spinners indicate a sum of exactly 5’
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There are two locations on the axes where one of the spinners lands on a 3 and the sum is 5. Therefore;
Note there are 16 points on the axes.
We need to find P(B|A); we will need to use the following formula;
We’re finding P(sum of 5 | at least one 3);
We have already found the P(sum of 5 and at least one 3). We can use the axes again to find P(at least one 3). There are 7 outcomes which contain at least one 3, these have been shaded yellow in the diagram below;
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Therefore P(at least one 3) is 7/16 therefore;
A and B are two events such that P(A) = 0.2, P(B) = 0.6 and P(A|B) = 0.3
Find
- P(B|A)
- P(A’ n B’)
- P(A’ n B)
Drawing the Venn diagram first will simplify the process of solving the question; We will need to find the intersection first. We do that by using the multiplication rule;
Starting with the intersection;
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-
P(A|B) = P(B n A)/P(A)= 0.18/0.2 = 0.9
- P(A’ n B) is the intersection of ‘not A’ and ‘not B’
P(A’ n B’) = 0.38
-
P(A’ n B) is the intersection of ‘not C’ with B.
P(A’ n B) = 0.42
For the two events A and B, P(A) = 3/10 P(B) = 2/5 and P(B u A’) = 1/2
Find;
- P(B|A)
- P(B|A’)
- P(A)P(B|A) + P(A’)P(B|A’)
- We will need to use the addition rule first to work out the intersection;
P(B n A) = P(B) + P(A) – P(B u A)= 2/5+3/10–1/2 = 1/5
Now we can use the following formula to find the conditional probability;
P(B|A) = P(B n A)/P(A)= 1/53/10 = 2/3 - Drawing Venn diagram will be useful here;
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The probability of B overlapping with ‘not A’ is B without the intersection. This has been shown on the Venn diagram below;
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So; P(B n A’) = 1/5
We know that P(A’) is simply 1 – P(A) = 1 – 3/10, therefore;P(B|A’) = P(A n B’)/P(A)‘
1/5/1- 3/10 = 2/7 -
P(A)P(B|A) + P(A’)P(B|A’)
Here we can just use the values we have found so far in the previous answers and the Venn diagrams.
3/10 × 2/3 + 7/10 × 2/7 = 2/5This is the same as the probability of event B illustrated in the Venn diagram below;
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…where P(B n A) = P(A)P(B|A) and P(A’)P(B|A) = P(B n A’)
