# Conditional Probability

From a Venn diagram, it is easy to see the relative probabilities of an event A, B, both A and B, or neither A nor B.

This is assuming a single, random occurrence. But, what effect does knowing extra information have on the probability of the events?

Take for example, the case of a school year with 66 pupils. 22 study German, 15 study French, and 37 do not take any language. 8 pupils take both languages.

If a child is selected at random, the chances he or she studies French are \$P(F) = {n(F)}/{n(U)} = {15}/{66} = 0.23\$.

If we know the child takes German (B), what is the probability she also takes French (A)?

\$P(A|B)\$ means 'what is the probability of A given B?'.

\$P(A|B) = {n(A∩B)}/{n(B)} = {8}/{22} = 4/{11} = 0.36\$.

This is the conditional probability, since the outcome of A is dependent on the outcome of B.

### Independent Events

If event A is independent of event B, \$P(A) = P(A|B) = P(A|B')\$. The probability of A occurring does not change as a result of the other event occurring: \$P(A∩B) = P(A) × P(B)\$.

Of the 38 million coffee drinkers in a country, 24 million take milk, 16 million take sugar, and 10 million take neither milk nor sugar.

(a) How many drink their coffee with both milk and sugar?

A lady enters a cafeteria. Find the probability that:

(b) She adds milk but not sugar.

(d) If she takes sugar, she does not add milk.

#### Solution

 a) Let \$n(A ∩ B) = x\$ \$(24 - x) + x + (16 - x) + 10 = 38\$ \$x = 12\$ Therefore \$P(A ∩ B) = {12}/{38} = 6/{19}\$ (b) \$P(A∩B') = {24-12}/{38} = {6}/{19}\$ (c) \$P(B|A) = {P(B∩A)}/{P(A)} = {{12}/{38}}/{{24}/{38}} = {12}/{24} = 1/2\$ (d) \$P(A'|B) = {P(A'∩B)}/{P(B)} = {{4}/{38}}/{{16}/{38}} = {4}/{16} = 1/4\$

## Probability Tree Diagrams

Probability tree diagrams break probabilities of a series of events into different independent paths

When more than one event occurs, tree diagrams can create a division of event paths, which makes it easy to visualise the distributions of probabilities for each part of the system.

### With Replacement

If the original conditions are reinstated before a repeat of the trial, as in the case of tossing a coin, or drawing a card, then replacing it, then the probability of a particular sequence of events (results of the trial) is the product of the probabilities of each step in the path.

### Without Replacement

Without replacement, each subsequent stage along the path has different conditions, so the probabilities change accordingly.

### Example

There are equal numbers of boys and girls in a school. One in ten boys and one in ten girls walk to school each day. One in three boys, and one out of two girls are brought by car. The remaining children come by bus.

a) What proportion of all children in the school are girls who come by bus?

b) What proportion of the children come by bus?

### Solution

a) \$P(G+b) = P(G) × P(b) = 0.5 × 0.4 = 0.2 = 1/5\$

20% of girls come by bus.

b) \$P(\$bus\$) = P(G+b) + P(B+b) = 0.2 + 0.285 = 0.485\$

48.5% of all children come by bus.

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