Independent Events

Events that do not affect each other

Roll a fair dice and toss a fair coin. Whatever the dice shows, the coin still lands heads with probability $\frac{1}{2}$1/2: the dice cannot change the coin. Two events are independent when one happening does not change the chance of the other.

On a tree diagram, independence means the second branches are the same after every first branch. After a six, a head has chance $\frac{1}{2}$1/2; after any other face, a head still has chance $\frac{1}{2}$1/2.

The dice first, a six or one of the faces 1 to 5, then the coin. Both pairs of coin branches read $\frac{1}{2}$1/2 and $\frac{1}{2}$1/2, because the dice does not change the coin. The path six then head is colored.

Every pair

To see what the chance of a six and a head is, list every pair. Each of the 6 faces can come with each of the 2 sides of the coin, so there are 6 × 2 = 12 pairs, from 1 with a head to 6 with a tail.

The 6 faces are equally likely and the 2 sides are equally likely, and neither affects the other, so all 12 pairs are equally likely. Each has probability $\frac{1}{12}$1/12.

The coin down the side, H for a head and T for a tail, and the dice across the top: 2 rows of 6, one cell for each of the 12 pairs.

Count the pair

Only one of the 12 cells is a six with a head, the cell 6H. So the probability of a six and a head is $\frac{1}{12}$1/12.

The one cell that is a six with a head: 1 of the 12 pairs.

Why the chances multiply

The grid has 12 cells because it has 6 columns and 2 rows: 12 = 6 × 2. So $\frac{1}{12}=\frac{1}{6}\times \frac{1}{2}$1/12 = 1/6 × 1/2. The chance of a six and a head is the chance of a six times the chance of a head.

The same holds for any event on the dice with any event on the coin. An even number and a head fill 3 cells of the head row, so the chance is $\frac{3}{12}=\frac{1}{4}$3/12 = 1/4, and $\frac{3}{6}\times \frac{1}{2}=\frac{3}{12}$3/6 × 1/2 = 3/12 as well. For independent events A and B, P(A and B) = P(A) × P(B).

An even number and a head: the cells 2H, 4H and 6H, 3 of the 12. That is 3 of the 6 columns in 1 of the 2 rows, so $\frac{3}{6}\times \frac{1}{2}$3/6 × 1/2.

A block on a grid

Roll a red dice and a blue dice. What is the probability that the red dice shows 1 or 2 and the blue dice shows 1, 2 or 3? The cells that win form a block 2 rows tall and 3 columns wide, so 2 × 3 = 6 of the 36 cells, and the probability is $\frac{6}{36}=\frac{1}{6}$6/36 = 1/6.

Multiplying the chances gives the same: $\frac{2}{6}\times \frac{3}{6}=\frac{6}{36}$2/6 × 3/6 = 6/36. The block's rows are the red dice's chance and its columns are the blue dice's chance, and the number of cells is rows times columns.

The red dice down the side and the blue dice across the top. Red 1 or 2 with blue 1, 2 or 3 is a block of 2 rows by 3 columns: 6 of the 36 cells.

Events that are not independent

Two events from one roll of a dice can depend on each other. Take "rolls an even number", the faces 2, 4 and 6, and "rolls more than 3", the faces 4, 5 and 6. Each has probability $\frac{1}{2}$1/2, so if they were independent, the chance of both would be $\frac{1}{2}\times \frac{1}{2}=\frac{1}{4}$1/2 × 1/2 = 1/4.

Count instead: the faces 4 and 6 are in both events, so the chance is $\frac{2}{6}=\frac{1}{3}$2/6 = 1/3, not $\frac{1}{4}$1/4. Knowing the roll is more than 3 leaves the faces 4, 5 and 6, and 2 of those 3 are even, so the chance of even has changed from $\frac{1}{2}$1/2 to $\frac{2}{3}$2/3. The events are not independent, and their chances do not multiply.

Two cards drawn without putting the first back are not independent either: the first card changes what is left for the second.

The usual mistakes

Adding the chances: $\frac{1}{6}+\frac{1}{2}=\frac{4}{6}$1/6 + 1/2 = 4/6. Both events must happen, which is less likely than either one alone, so the answer must be smaller than $\frac{1}{6}$1/6, and $\frac{1}{12}$1/12 is.

Adding the denominators, as in $\frac{1}{\left(6+2\right)}=\frac{1}{8}$1/(6 + 2) = 1/8, or adding the tops and the bottoms, $\frac{2}{8}$2/8. Taking $\frac{1}{2}$1/2 of $\frac{1}{6}$1/6 cuts each sixth in half, so there are 6 × 2 = 12 equal parts, not 8.

Multiplying the chances of events that are not independent, such as even and more than 3 on one roll. Check first that one event does not change the chance of the other.

Two lifts

In the application below, two lifts in an office block break down independently, with chances $\frac{1}{5}$1/5 and $\frac{1}{4}$1/4 on any day. In part (a), the chance that both are out of order is the product, $\frac{1}{5}\times \frac{1}{4}=\frac{1}{20}$1/5 × 1/4 = 1/20.