###### Define independence and calculate prob ...

Define independence and calculate probabilities of independent events 1.1 Defining Independence In probability... **Read More**

Before we move on to Bayes Theorem, we need to learn about the **law of total probability.**

The law of total probability states that if E is an event, and \(A_1, A_2, \cdots A_n\) are the partition of the sample space, then

$$P(E)=P(A_1 \cap E)+P(A_2 \cap E)+\cdots P(A_n \cap E)$$

We can use the law intuitively by recalling that:

$$A=(A\cap B)\cup (A\cap B^c)$$

Consider the following Venn Diagram:

Note that \(B^c\) (also be written as \(B’\)) is the complement of an event B.

Therefore,

$$P(A)=P(A\cap B)+P(A\cap B^c)$$

Recall the conditional probability that:

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

So that:

$$P(A\cap B)=P(B)\bullet P(A|B)$$

Similarly,

$$P(A\cap B^c)=P(B^c)\bullet P(A|B^c)$$

Thus,

$$ P\left( A \right) =P\left( A|B \right) \bullet P\left( B \right) +P\left(A|B^c \right)\bullet \left( P\left( B^c \right) \right) $$

At this point, we can write the expression for \(P(B|A)\) as:

$$P(B|A)=\frac{P(A\cap B)}{P(A)}=\frac{P(A)\bullet P(A|B)}{P\left( A|B \right) \bullet P\left( B \right) +P\left(A|B^c \right)\bullet \left( P\left( B^c \right) \right)}$$

The last expression is referred to as **Bayes’ Theorem. **

Similarly,

$$P(B^c |A)=\frac{P(A\cap B^c)}{P(A)}=\frac{P(A)\bullet P(A|B^c)}{P\left( A|B \right) \bullet P\left( B \right) +P\left(A|B^c \right)\bullet \left( 1-P\left( B \right) \right)}$$

If \(E\) is an event and \(A_1 , A_2 ,…, A_3\) are the partition of a sample space, then:

$$\begin{align} P(A_i |E)&=\frac{A_i \cap E}{P(E)}\\ &=\frac{P(A_i)\bullet P(E|A_i)}{P(A_1)\bullet P(E|A_1)+P(A_2)\bullet P(E|A_2)+\cdots +P(A_n)\bullet P(E|A_n)} \end{align}$$

Consider the following examples:

In a given population of students, 25% play baseball, and 30% play basketball. Also, the probability of a student playing baseball given that they also play basketball is 10%.

Calculate the probability that a student plays baseball given that they do not play basketball.

**Solution**

Let event \(A\) be “plays baseball” and event \(B\) be “plays basketball.” Using the law of total probability, we have:

$$ \begin{align} P\left( A \right) & =P\left( A|B \right) \bullet P\left( B \right) +P\left(A|B^c\right)\bullet \left( 1-P\left( B \right) \right) \\ 0.25 &= 0.10 \bullet 0.30 + x \left(1-0.30 \right) \\ 0.22 &= x \left(0.70 \right) \\ x &= \frac{0.22}{0.70} = 0.314 \ or \ 31.4\% \end{align} $$

An insurance company deals with three insurance policies: 40% of life insurance, 25% of car insurance, and 35% of health insurance. The probability that a life insurance policyholder will file a claim in a given year is 0.50. The probability that a car insurance policyholder will file a claim in a given year is 0.20. Lastly, the probability that a health insurance policyholder will file a claim in a given year is 0.10.

At the course of this year, a policyholder files a claim.

Calculate the probability that the claim comes from the car insurance policyholder.

**Solution**

Let:

\(L\) = proportion of life insurance policyholders.

\(R\) = proportion of car insurance policyholders.

\(H\) = proportion of health insurance policyholders.

\(C\) = event that a claim is made.

From the information given in the question, we have:

$$ P(L) = 0.40,\ P(R) = 0.25,\ P(H) = 0.35 $$

Also, we have:

$$ P(C|L) = 0.50,\ P(C|R) = 0.20,\ P(C|H) = 0.10 $$

We need, \(P(R|C)\). Using the Bayes’ theorem, we know that:

$$\begin{align}P(R|C)&=\frac{P(R)\bullet P(C|R)} {P(R)\bullet P(C|R)+P(L)\bullet P(C|L)+P(H)\bullet P(C|H)}\\ &=\frac{0.25\times 0.20}{0.25\times 0.20+0.40\times 0.50 +0.35\times 0.10}\\ &=0.17543 \approx 17.54\%\end{align}$$

**Learning Outcome**

**Topic 1.g: General Probability – State Bayes Theorem and use it to calculate conditional probabilities.**