Multiplying Along a Tree Without Replacement

One path through the tree

The bag holds 3 red counters and 2 blue, and two are drawn without replacement. Drawing red and then red again is one path through the tree: the branch red, $\frac{3}{5}$3/5, then the branch red, $\frac{2}{4}$2/4.

R for a red counter and B for a blue one. The path red then red is colored: $\frac{3}{5}$3/5 on the first branch and $\frac{2}{4}$2/4 on the second.

Multiply along the path

Of all pairs of draws, $\frac{3}{5}$3/5 start with red. Of those, $\frac{2}{4}$2/4 go on to draw red again. So red then red is $\frac{2}{4}$2/4 of $\frac{3}{5}$3/5: $\frac{3}{5}\times \frac{2}{4}=\frac{6}{20}=\frac{3}{10}$3/5 × 2/4 = 6/20 = 3/10.

Count to check. Name the counters R1, R2, R3, B1 and B2. There are 5 × 4 = 20 equally likely ordered pairs of two different counters, which is where the denominator 20 comes from. Red then red takes one of 3 reds first and one of the other 2 reds second: 3 × 2 = 6 pairs, so $\frac{6}{20}$6/20.

The first counter down the side and the second across the top, with the same counter twice crossed out. The 6 colored cells are red then red: 6 of 20.

One of each color: two paths

One red and one blue can happen in two orders. Red then blue is $\frac{3}{5}\times \frac{2}{4}=\frac{6}{20}=\frac{3}{10}$3/5 × 2/4 = 6/20 = 3/10: after a red, 2 of the 4 counters left are blue.

Blue then red is $\frac{2}{5}\times \frac{3}{4}=\frac{6}{20}=\frac{3}{10}$2/5 × 3/4 = 6/20 = 3/10: after a blue, 3 of the 4 counters left are red. The two paths have the same chance here, but they are worked out from different branches.

Each path ends on the product of its two branches. Blue then red is colored, $\frac{2}{5}\times \frac{3}{4}=\frac{6}{20}$2/5 × 3/4 = 6/20; red then blue, the second path, is $\frac{3}{5}\times \frac{2}{4}=\frac{6}{20}$3/5 × 2/4 = 6/20.

Add the paths

A pair of draws follows exactly one path, so the paths are mutually exclusive, and the chance of one of each color is the sum: $\frac{3}{10}+\frac{3}{10}=\frac{6}{10}=\frac{3}{5}$3/10 + 3/10 = 6/10 = 3/5.

On the grid, the cells with one red and one blue are the 6 with red first and blue second and the 6 with blue first and red second: 12 of 20, which is $\frac{3}{5}$3/5.

One of each color: the 6 cells of red then blue and the 6 cells of blue then red, 12 of the 20.

Every path adds to 1

Blue then blue is $\frac{2}{5}\times \frac{1}{4}=\frac{2}{20}$2/5 × 1/4 = 2/20. The four paths add to $\frac{6}{20}+\frac{6}{20}+\frac{6}{20}+\frac{2}{20}=\frac{20}{20}=1$6/20 + 6/20 + 6/20 + 2/20 = 20/20 = 1, because two draws must end on one of them.

The same color is red then red or blue then blue: $\frac{6}{20}+\frac{2}{20}=\frac{8}{20}=\frac{2}{5}$6/20 + 2/20 = 8/20 = 2/5. One of each and the same color are complements, and $\frac{3}{5}+\frac{2}{5}=1$3/5 + 2/5 = 1.

A bag of 2 red and 4 blue

Two counters are drawn without replacement from 2 red and 4 blue. Red then red is $\frac{2}{6}\times \frac{1}{5}=\frac{2}{30}=\frac{1}{15}$2/6 × 1/5 = 2/30 = 1/15: after a red, 1 red is left among 5 counters.

One of each is red then blue, $\frac{2}{6}\times \frac{4}{5}=\frac{8}{30}$2/6 × 4/5 = 8/30, plus blue then red, $\frac{4}{6}\times \frac{2}{5}=\frac{8}{30}$4/6 × 2/5 = 8/30, which makes $\frac{16}{30}=\frac{8}{15}$16/30 = 8/15.

The usual mistakes

Keeping the second denominator at the first total: $\frac{2}{6}\times \frac{1}{6}=\frac{2}{36}$2/6 × 1/6 = 2/36. After one draw only 5 counters are left.

Keeping the drawn red in the second numerator: $\frac{2}{6}\times \frac{2}{5}=\frac{4}{30}$2/6 × 2/5 = 4/30. One red has already left the bag, so 1 red is left.

Stopping at one path for one of each, $\frac{8}{30}$8/30. Blue then red also gives one of each, and its chance adds on.

Adding along a path or multiplying across paths. Multiply along a path, because both draws happen; add different paths, because only one of them happens.

Faulty bolts

In the application below, a bolt comes from one of two machines and may be faulty. Part (a) multiplies along the two paths that end in a faulty bolt and adds them; part (b) asks which machine a faulty bolt came from.