Conditional probability without replacement

Without replacement means an item you pick isn’t put back before the next pick. So after the first pick, one fewer item remains. If that first item was a target, one fewer target remains too. The sample space, the set of items you’re picking from, shrinks with every pick.

Say a case holds $5$ marked cards and $3$ unmarked cards, and you pick two without replacement. The probability that both are marked is

The first fraction uses all $8$ cards. By the second pick, one marked card is gone, so $4$ marked cards remain out of $7$.

Multiplying the chances one pick at a time like this is a sequential product. It’s quick here because the target happens in one simple order: marked, then marked. If the target can happen in several orders, or the question adds a condition about the whole pair, counting whole pairs is often shorter.

Often a question cares which two items you end up with, not which one came first. Then you count each pair once.

The notation

is read “$n$ choose $2$.” It’s the number of unordered pairs you can make from $n$ items:

For example,

Why divide by $2$? There are $8$ choices for a first item and $7$ for a second, but that counts every pair twice. Picking $A$ then $B$ gives the same pair as picking $B$ then $A$.

With two groups, say $a$ target items and $b$ other items, every pair is one of three types:

The middle count is $ab$ because any of the $a$ targets can pair with any of the $b$ others. The three types cover every pair, so they add up to the total:

You don’t need to memorize a pile of formulas. Name the pair type, decide which group fills each of its two spots, and count each pair once.

At least one target covers more than one case. In a pair, it means exactly one target or two targets. It’s usually quicker to count the one case it leaves out and subtract:

With $a$ target items and $b$ other items, that’s

When the condition says the pair has at least one target, this count is your denominator.

The same idea works for a group of three. The notation $\binom{n}{3}$ counts the unordered groups of $3$ you can choose from $n$ items:

This time you divide by $3\cdot2\cdot1=6$, because each group of three can be picked in $6$ different orders.

Say $12$ reports include $4$ ecology reports and $8$ geology reports, and you pick three without replacement. Given that at least one is an ecology report, what’s the probability that exactly one is?

Start with the condition. Subtract the all-geology groups from all the groups:

Now count the target inside it. For exactly one ecology report, choose $1$ of the $4$ ecology reports and $2$ of the $8$ geology reports:

So the probability is

The numbers get bigger with three picks, but the steps are the same.

Two methods work here, and the question tells you which one is shorter.

• Use a sequential product when the target happens in one short order and no condition covers the whole group. Two marked cards in a row, $\frac{5}{8}\cdot\frac{4}{7}$, is a good example.
• Use pair or small-group counts when order doesn’t matter or the target can happen in more than one order. One priority packet and one standard packet could come in either order, and a pair count covers both at once.
• Also use counts when the condition is about the finished selection, especially at least one. That’s how the packet example found its $135$ pairs.

Do the counting by hand. Deciding which selections fit the condition is the real question, and a calculator can’t decide that for you. Once your counts are set, it’s fine to use one to reduce a fraction.