Probability Without Replacement
Grade 6What changes when you don't replace?
In a without replacement problem, each item picked stays out of the bag. This means two things change for the second pick: the total goes down by 1, and the number of that colour also goes down by 1 (if you picked one of them).
This is conditional probability in action — what you pick first directly affects the probability of what comes next.
Worked example
EXAMPLE
A bag contains 5 red and 3 blue counters. Two counters are picked without replacement. Find P(both red).
First pick: P(red) = 5/8.
After one red is removed, there are now 4 red and 3 blue left — 7 counters total.
Second pick (given first was red): P(red) = 4/7.
P(both red) = 5/8 × 4/7 = 20/56 = 5/14.
Drawing the tree diagram
Draw the tree as normal, but write different probabilities on the second set of branches depending on which outcome the first branch took. If the first pick was red, the second-branch probabilities use a reduced red count. If the first pick was blue, the red count stays the same but blue reduces.
After the first pick, the denominator is always one less than before — write that new total on the second set of branches before you fill in the numerators.
Worked example — at least one blue
EXAMPLE
Using the same bag (5 red, 3 blue), find P(at least one blue counter in two picks, without replacement).
"At least one blue" is easier to find using the complement: P(at least one blue) = 1 − P(no blue) = 1 − P(both red).
We already know P(both red) = 5/14.
P(at least one blue) = 1 − 5/14 = 9/14.
