Conditional Probability

Probability

Counting inside the condition

With a listed or tabulated sample space, cross out every outcome the condition excludes
What remains is the new total, and the successes are counted from those survivors only
Working from a completed grid makes this reliable, since the condition simply selects some of the cells

Exam tip

The condition replaces the total, so never divide by the overall number: showing the count as a fraction of the condition's group, such as 9/22 rather than 9/40, is what earns the mark. Order matters — P(B given A) is not P(A given B). The syllabus states outright that the notation P(A|B) and any conditional-probability formula are not required, so read the answer off the diagram.

Worked example

A condition applied to a sample space

One spinner is numbered 1, 2, 3 and another is numbered 4, 5, 6. Both are spun and the numbers are added. Given that the total is odd, find the probability that one of the numbers spun is 3.

Solution:

There are 3 × 3 = 9 equally likely outcomes altogether
The totals are 5, 6, 7 from a first spin of 1; 6, 7, 8 from a first spin of 2; and 7, 8, 9 from a first spin of 3
The odd totals come from (1, 4), (1, 6), (2, 5), (3, 4) and (3, 6), so 5 outcomes meet the condition
The 9 is not the total any more: the restricted total is 5
Of those 5 outcomes, the ones including a 3 are (3, 4) and (3, 6), so 2 succeed
P(one number is 3, given the total is odd) = 2/5