There are various rules in probability that can be used to calculate the probability of several events. Multiplication rule is used to find the probability of an intersection of two events. It make sense that the definition of conditional probability involves an intersection because conditional probability is where the multiplication rule comes from. The probability that both $C$ and $D$ occur when the experiment is performed is $P(C\ \cap\ D)$ = $P(C)P(C|D)$ or $P(C\ \cap\ D)$ = $P(D)P(D|C)$

What is the conditional probability that a family with two children has two girls, given that they have at least one girl?

The all possibilities can be denoted by the set $\{ BB, BG, GB, GG \}$. The first letter represents the sex of the first child, while the second denotes the sex of the second child. The sample space has four outcomes.

The set of outcomes for having two girls is the subset of the set of outcomes for having at least one girl.

P(two girls| at least one girl) = $\frac{P(Two\ girls\ n at\ least\ one\ girl)}{P(at\ least\ one\ girl)}$

= $\frac{\frac{1}{4}}{\frac{3}{4}}$

= $\frac{1}{3}$

$\Rightarrow$ P(two girls| at least one girl) = $\frac{1}{3}$.

Two math workshops were organized by the Math club of the school. The first workshop was attended by $40 \%$ of High school students. Both the workshops were attended by $25 \%$ of High School students. What percent of those who attended the first workshop also took part in the second?

The first workshop was attended by $40 \%$ of High school students and both the workshops were attended by $25 \%$ of High School students.

Percent of those students who attended the first workshop as well as second = ?

P(attended the second | attended the first) = $\frac{P(attended\ both)}{P(attended\ the\ first)}$

= $\frac{0.25}{0.4}$

= $0.625$ = $62.5$%

Hence $62.5 \%$ of the students who attended the first workshop also took part in the second.

Class | Male | Female | Total |
---|---|---|---|

Freshman | 1265 | 1135 | 2400 |

Sophomore | 1235 | 965 | 2200 |

Total | 2500 | 2100 | 4600 |

A TV channel selects one of the students to participate in their program on Student’s political preferences. Using the table let us try to answer the following questions.

These two are instances of conditional probabilities:

The fact that the selected student is a freshman, the sample space is restricted to all students in the freshman class. So the number of female against the row is to be compared to the total number of students in freshman class.

**Step 1:**

**Step 2:**

**Step 3:**

**Step 4:**

$P(female|freshman)$ = $\frac{Number\ of\ females\ in\ the\ freshman\ class}{Number\ of\ students\ in\ freshman\ class}$ = $\frac{1135}{2400}$ $˜0.4729$

Dividing both the numerator and denominator by Total number of students,

$P(female|freshman)$ = $\frac{\frac{Number\ of females\ in\ freshman\ class}{Total\ number\ of\ students}}{\frac{Number\ of\ students\ in\ freshman\ class}{Total\ number\ of\ students}}$

Now the top and bottom can be replaced in terms of probabilities as

$P(female|freshman)$ = $\frac{P(femalenfreshman)}{P(freshman)}$ …………………………..(1)

Now for the second question, the sample space is restricted to all males in the University. The number of sophomores under the male column has to be compared with the total number of males.

$P(sophomore|male)$ = $\frac{Number\ of\ males\ in\ the\ sophomore\ class}{Total\ number\ of\ males}$

= $\frac{1265}{2500}$

= $0.506$

Dividing the numerator and denominator by the total number of males,

$P(Sophomore|Male)$ = $\frac{\frac{Number\ of\ males\ in\ sophomore\ class}{Total\ number\ of\ students}}{\frac{Total\ number\ of\ males}{Total\ number\ of\ students}}$

Replacing the numerator and denominator with probabilities,

$P(Sophomore|Male)$ = $\frac{P(Sophomoren\ Male)}{P(Male)}$ ………………………………(2)

Observing the two equations we can arrive at the general formula for conditional probability.