With and Without Replacement

A bag holds three red counters and two blue. You take one, note its colour, and put it back before you take another. That is called drawing with replacement. Because the counter goes back, the bag is exactly the same for the second draw, so the second set of branches is the same as the first: three fifths for red and two fifths for blue. Multiply along each path. Red then red is three fifths times three fifths, which is nine over twenty five.

Now the first counter is not put back. That is drawing without replacement. After the first draw there are only four counters left, so every second-draw fraction is out of four. If the first was red, two red and two blue are left, so red is now two quarters. If the first was blue, three red and one blue are left, so red is three quarters and blue is one quarter. Multiply along the paths as before. Red then red is three fifths times two quarters, which is six twentieths.

Two counters of the same colour means two reds or two blues. From the tree without replacement, red then red is six twentieths and blue then blue is two twentieths. The event can happen either way, so add the paths: eight twentieths, which simplifies to two fifths.

One of each colour also happens on two paths: red then blue, and blue then red. Each is six twentieths, so the total is twelve twentieths, or three fifths. Notice that the order you draw them in creates two separate paths, which is why one of each is so often doubled. There is also a quick check: one of each is everything that is not two the same, and one minus two fifths is three fifths.

Put the two methods side by side. With replacement, two reds is nine over twenty five; without, it is three tenths, which is smaller, because once a red is gone, red is rarer. For the same reason, one of each is more likely without replacement: three fifths, against twelve over twenty five. If a question says two counters are taken at once, or chosen together, treat it as without replacement. Quiz and exercises now.