The probability tree and the law of total probability

In a probability tree each stage of the experiment is a column of branches. The probability on a branch in the second stage is a conditional probability: the probability of that outcome given the path that led to it. So after each outcome of the first stage the probabilities can differ.

At a school $\displaystyle \frac{2}{5}$ of the students are boys.
20% of the boys are active in a youth movement.
30% of the girls are active in a youth movement.
A student is chosen at random.

• The first stage: boy or girl – $\displaystyle \frac{2}{5}$ and $\displaystyle \frac{3}{5}$.
• The second stage after “boy”: active $\displaystyle \frac{1}{5}$ – this is $\displaystyle P(A|M)$, and not active $\displaystyle \frac{4}{5}$.
• The second stage after “girl”: active $\displaystyle \frac{3}{10}$ – this is $\displaystyle P(A|F)$, and not active $\displaystyle \frac{7}{10}$.
• The branches leaving one point are all the possibilities there, so their probabilities add up to 1.

A path in the tree is an intersection: “a boy and active”. Its probability is the product of the probabilities on its branches – exactly the formula of the last lesson, $\displaystyle {P(A\cap B)}={P(B)\cdot P(A|B)}$:

• An active boy: $\displaystyle {\frac{2}{5}\cdot\frac{1}{5}}={\frac{2}{25}}={0.08}$.
• An active girl: $\displaystyle {\frac{3}{5}\cdot\frac{3}{10}}={\frac{9}{50}}={0.18}$.

The probabilities of all the paths add up to 1: every student is on exactly one path.

What is the probability that a student chosen at random is active in a youth movement? The event “active” is made of two disjoint paths – an active boy and an active girl – so they are added:

This is the law of total probability for two events:

The law is a weighted mean of the conditional probabilities: 20% and 30%, with the weights $\displaystyle \frac{2}{5}$ and $\displaystyle \frac{3}{5}$. So the result, 26%, lies between 20% and 30%, and nearer to 30% – there are more girls.

In the area model the whole square is 1. The boys’ strip is $\displaystyle \frac{2}{5}$ wide, and the girls’ $\displaystyle \frac{3}{5}$; each strip is cut by the probabilities after it:

The marked area is all the active students: a low rectangle in the narrow strip and a taller one in the wide strip, $\displaystyle {0.08+0.18}={0.26}$.

In an experiment of several stages the tree goes on with a column for each stage, and the probability on a branch can depend on the whole path before it.

A basketball player takes three free throws, one after another.
On the first throw she scores with probability $\displaystyle 0.7$.
After a score she scores on the next throw with probability $\displaystyle 0.8$.
After a miss she scores on the next throw with probability $\displaystyle 0.5$.

The paths of “exactly two scores” are marked:

• Score, score, miss: $\displaystyle {0.7\cdot0.8\cdot0.2}={0.112}$.
• Score, miss, score: $\displaystyle {0.7\cdot0.2\cdot0.5}={0.07}$.
• Miss, score, score: $\displaystyle {0.3\cdot0.5\cdot0.8}={0.12}$.

The probability of exactly two scores: $\displaystyle {0.112+0.07+0.12}={0.302}$.

A factory has two machines.
Machine A makes 60% of the parts, and machine B – 40%.
5% of machine A’s parts are defective, and 10% of machine B’s.
A part is chosen at random from all the factory’s parts.

Write the probabilities on the branches and each path’s probability, and mark the paths of “the part is defective”.

What is the probability that the part was made by machine A and is defective?

What is the probability that the part is defective?

Explain why the branch “defective” after “machine B” says $\displaystyle 0.1$, and not $\displaystyle 0.4\cdot0.1$.

Three boxes, $\displaystyle A$, $\displaystyle B$ and $\displaystyle C$, hold black and white balls.
Box $\displaystyle A$ holds two black balls and three white ones.
Box $\displaystyle B$ holds three black balls and two white ones.
Box $\displaystyle C$ holds four black balls and one white one.
A box is chosen at random, and one ball is taken from it at random.

Complete the tree, and work out the probability of taking a white ball.

The probability that it rains today is $\displaystyle 0.3$.
If it rains today, the probability that it rains tomorrow too is $\displaystyle 0.6$.
If it does not rain today, the probability that it rains tomorrow is $\displaystyle 0.2$.

What is the probability that it rains tomorrow?

Yossi plays three games of backgammon, one after another. In each game he wins or loses (no draws).
In the first game he wins with probability $\displaystyle 0.7$.
If Yossi won a game, the probability that he wins the next one is $\displaystyle 0.6$.
If he lost a game, the probability that he loses the next one too is $\displaystyle 0.6$.

Complete the tree.

What is the probability that Yossi wins all three games?

What is the probability that Yossi loses the second game and wins the third?