Conditional Probability

For two events $A$ and $B$,

$P(A \cap B) = P(A)\; .\; P(B \mid A), \quad P(A) >0$

$= P(B) \;.\; P(A \mid B), \quad P(B) >0$

where $P(B \mid A)$ represents conditional probability of occurrence of $B$ when the event $A$ has already happened and $P(A \mid B)$ is the conditional probability of happening of $A$, given that $B$ has already happened.

Proof

In the usual notations, we have

$P(A) = \frac{n(A)}{n(\Omega)}; \quad P(B) = \frac{n(B)}{n(\Omega)}\quad and \quad P(A \cap B) = \frac{n(A \cap B)}{n(\Omega)}$ -------- (1)

For the conditional event $A \mid B$, the favorable outcomes must be one of the sample points of $B$, that is, for the event $A \mid B$, the sample space is $B$ and out of the $n(B)$ sample points, $n(A \cap B)$ pertain to the occurrence of the event $A$. Hence

$P(A \mid B) = \frac{n(A \cap B)}{n(B)}$

$\Rightarrow \quad n(A \cap B) = n(B) \;. \;P(A \mid B)$

$\Rightarrow \quad \frac{n(A \cap B)}{n(\Omega)} = \frac{n(B)}{n(\Omega)}\; . \;P(A \mid B)$ [Divided on both sides by $n(\Omega)$]

$\Rightarrow \quad P(A \cap B) = P(B)\; . \; P(A \mid B).$

Similarly, we get from $(1)$

$P(A \cap B) = P(A)\; . \;P(B \mid A).$

Thus, we have proved that "the probability of the simultaneous occurrence of two events $A$ and $B$ is equal to the product of the probability of one of the events and the conditional probability of the other, given that the first one has occurred".

Students can get the online Probability calculator for help with solving probability problems.

Corollary 1

$P(B \mid A) = \frac{P(A \cap B)}{P(A)}$ and $P(A \mid B) = \frac{P(A \cap B)}{P(B)}$

Thus the conditional probabilities $P(B \mid A)$ and $P(A \mid B)$ are defined if, and only if, $P(A) \not= 0$ and $P(B) \not= 0$, respectively.

Corollary 2

$P(A \mid A) = 1$, for any event $A$.

Conditional probability Example 1

The probability that a construction job will be finished on time is $\frac{3}{5}$; the probability that there will be no strikes is $\frac{2}{3}$, and the probability that the job will be finished on time given that there are no strikes is $\frac{4}{5}$. Then

1. What is the probability that the job will be finished on time and there will be no strikes?

2. What is the probability that there will have been no strikes given that the job is finished on time?

Solution

Let

$A$ be the event that the construction job will be finished on time.

$B$ be the event that there will be no strikes.

Then,

$A \cap B$ : The event that the job will be finished on time and there will be no strikes.

$A \mid B$ : The event that the job will be finished on time given that there are no strikes.

$B \mid A$ : The event that there will have been no strikes given that the job is finished on time.

Given that

$P(A) = \frac{3}{5}, \quad P(B) = \frac{2}{3}, \quad and \quad P(A \mid B) = \frac{4}{5}.$

To find

1. $P(A \cap B),$

2. $P(B \mid A).$

1. From Multiplication Theorem of Probability, we have

$P(A \cap B) = P(B)\; . \; P(A \mid B)$

$= \frac{2}{3} \times \frac{4}{5}$

$= \frac{8}{15}.$

2. Again, from Multiplication Theorem of Probability, we have

$P(A \cap B) = P(A)\; . \; P(B \mid A)$

$\Rightarrow \quad P(B \mid A) = \frac{P(A \cap B)}{P(A)}$

$= \frac{\left(\frac{8}{15}\right)}{\left(\frac{3}{5}\right)}$

$= \frac{8}{15} \times \frac{5}{3}$

$ = \frac{8 \times 5}{15 \times 3}$

$= \frac{8}{3 \times 3}$

$= \frac{8}{9}.$

Conditional probability Example 2

It is known that $40\;\%$ of the students in a certain college are girls and $50\;\%$ of the students are above the median height. If $\frac{2}{3}$ of the boys are above the median height, what is the probability that a randomly selected student who is below the median height is a girl.

Solution

Let

$G$ be the event of selecting a girl,

$B (= \overline{G})$ be the event of selecting a boy,

$M$ be the event of selecting a student who is above the median height.

Given that $40\;\%$ of the students are girls. So, the remaining $60\;\%$ are boys. Thus, we have the probabilities

$P(G) = \frac{40}{100} = \frac{2}{5},$

$P(B) = \frac{60}{100} = \frac{3}{5}.$

Also given that $50\;\%$ of the students are above the median height. Therefore, the probability that the student selected is above the median height is

$P(M) = \frac{50}{100} = \frac{1}{2}.$

Then, the probability that the student selected is below the median height is given by

$P(\overline{M}) = 1 - P(M) = 1 - \frac{1}{2} = \frac{1}{2}.$

Also, given $\frac{2}{3}$ of the boys are above the median height. That is,

$P(M \mid B) = \frac{2}{3}.$

To find $P(G \mid \overline{M})$, the probability that a randomly selected student who is below the median height is a girl.

By Multiplication Theorem of Probability, we have

$P(M \cap B) = P(B) \; . \; P(M \mid B)$

$= \frac{3}{5} \times \frac{2}{3}$

$= \frac{2}{5}.$

Then, we have

$P(M \cap G) = P(M \cap \overline{B})$

$= P(M) - P(M \cap B)$

$= \frac{1}{2} - \frac{2}{5}$

$= \frac{5}{10} - \frac{4}{10}$

$= \frac{5 - 4}{10}$

$= \frac{1}{10}.$

Then, we have

$P(G \cap \overline{M}) = P(G) - P(G \cap M)$

$= \frac{2}{5} - \frac{1}{10}$

$= \frac{4}{10} - \frac{1}{10}$

$= \frac{4 - 1}{10}$

$= \frac{3}{10}.$

Now,

$P(G \mid \overline{M}) = \frac{P(G \cap \overline{M})}{P(\overline{M})}$

$= \frac{\left(\frac{3}{10}\right)}{\left(\frac{1}{2}\right)}$

$= \frac{3}{10} \times \frac{2}{1}$

$= \frac{3}{5}.$

Thus, the probability that a randomly selected student who is below the median height is a girl is $\frac{3}{5}.$

Students can also get help with Probability homework online.