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Compute conditional probability as the fraction of one event's outcomes that also belong to another event.

Compute conditional probability from a finite sample space by restricting to the given event

Problem
In \(S={1,2,3,4,5,6}\), restrict to outcomes greater than \(3\), then count the even outcomes within that conditioned set to find \(P(\text{even}|\text{greater}~\text{than}~3)\).
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Hint

Start with the condition “greater than 3,” then keep only those outcomes.

For \(P(A|B)\), restrict the sample space to \(B\) first, then count how many of those outcomes are also in \(A\).

Solution walkthrough

01

Restrict to the condition

\[\text{greater}~\text{than}~3~->~{4,5,6}\]

Conditional probability discards outcomes that do not satisfy the condition before counting the target event.

02

Count the conditioned sample space

\[\text{there}~\text{are}~3~\text{outcomes}\]

The new denominator is 3, not the original sample-space size 6.

03

Find even outcomes within it

\[{4,6},~\text{so}~\text{favorable}~\text{count}=2\]

Four and six are even; five is not.

04

Form and interpret

\[P(\text{even}|\text{greater}~\text{than}~3)=2/3\]

Two of the three outcomes in the restricted set are even.

+

Another way

  1. Use intersection 4,6 over condition set 4,5,6.

!

Common mistake

Do not use all six outcomes in the denominator after conditioning.

Solution walkthrough video