Conditional probability

Which group did you choose from?

• A student is chosen at random from the French group. What is the chance they also study German? Only French students can be chosen, so the total has changed.
• Conditional probability 条件概率 means the chance of an event given some information. This is an Extended topic. Use tables, Venn diagrams and trees; a special formula is not required.

A table makes the selected group visible

• In this class, 18 of the 30 students study French. Of these 18, 7 study German. A student is chosen at random from those who study French.
• Read across the French row: 7 of its 18 students meet the target. The two-way table 双向表 also shows the other students, but they cannot be chosen under this condition.

Reversing the condition changes the question

• Now choose a student at random from the German group. There are 12 German students, and 7 also study French, so the probability of French given German is $\dfrac{7}{12}$.
• Compare $\dfrac{7}{12}$ with $\dfrac{7}{18}$. Saying which group is selected matters. A condition can be known information about a person; it need not be an earlier event in time.

Read a Venn diagram by counting regions

• The same class has 11 students in French only, 7 in both groups, 5 in German only and 7 in neither. Given French, the selected region contains $11+7=18$ students.
• Given not French, use the region outside the French circle. It contains $5+7=12$ students, of whom 5 study German, so the probability is $\dfrac{5}{12}$.

The counts describe the same students as the table. Each student belongs to exactly one of the four regions.

On a tree, read the branch after the condition

• A bag contains 3 red and 2 blue counters. Choose two counters at random without replacement. If the first was blue, 3 red and 1 blue remain, so the chance of a second red is $\dfrac{3}{4}$.
• The blue-then-red path has probability $\dfrac{2}{5}\times\dfrac{3}{4}=\dfrac{3}{10}$. This is the chance of both results before either draw. Given the first is blue, use only its second-stage red branch: $\dfrac{3}{4}$.

A condition can select more than one path

• For the same bag, suppose you know the second counter is red, but the first colour is hidden. Two paths meet the condition: RR has probability $\dfrac{3}{5}\times\dfrac{2}{4}=\dfrac{6}{20}$, and BR has probability $\dfrac{2}{5}\times\dfrac{3}{4}=\dfrac{6}{20}$.
• Together these paths have probability $\dfrac{12}{20}$. The part that starts blue is $\dfrac{6}{20}$, so among the selected paths its share is $\dfrac{6}{20}\div\dfrac{12}{20}=\dfrac{1}{2}$. Count path probabilities, not just the number of paths; paths need not be equally likely.

Does knowing the first result change the chance?

• Before seeing the first colour, the second-red paths total $\dfrac{12}{20}=\dfrac{3}{5}$. Given the first was blue, the chance of a red second counter is $\dfrac{3}{4}$. The draws are dependent 相依的: knowing the first result changes the chance at the next stage.
• With replacement and a thorough mix, the chance of red on the second draw would stay $\dfrac{3}{5}$, whatever the first colour. Check the selection method before assuming that probabilities stay the same.

Explain the group and the target

• For a new problem, write down which people or outcomes can be chosen after using the given information. Then count which of them meet the target.
• Give the probability as a fraction, decimal or percentage, and explain your denominator. If you reverse the condition, work out the new total before calculating again.

Keep the condition in view