0.14 More probability  (Page 3/5)

We will soon discover that this is a statement of Bayes' formula .

Let us first visualize the problem.

We are given a sample space $S$ and two mutually exclusive events $JI$ and $J\text{II}$ . That is, the two events, $JI$ and $J\text{II}$ , divide the sample space into two parts such that $JI\cup J\text{II}=S$ . Furthermore, we are given an event $B$ that has elements in both $JI$ and $J\text{II}$ , as shown in the Venn diagram below.

From the Venn diagram, we can see that

$B=\left(B\cap JI\right)\cup \left(B\cap J\text{II}\right)$

and

$P\left(B\right)=P\left(B\cap JI\right)+P\left(B\cap J\text{II}\right)$

But the product rule in [link] gives us

$P\left(B\cap JI\right)=P\left(JI\right)\cdot P\left(B\mid JI\right)$       $P\left(B\cap J\text{II}\right)=P\left(J\text{II}\right)\cdot P\left(B\mid J\text{II}\right)$

Substituting in [link] , we get

$P\left(B\right)=P\left(JI\right)\cdot P\left(B\mid JI\right)+P\left(J\text{II}\right)\cdot P\left(B\mid J\text{II}\right)$

The conditional probability formula gives us

$P\left(JI\mid B\right)=\frac{P\left(JI\cap B\right)}{P\left(B\right)}$

Therefore,

$P\left(JI\mid B\right)=\frac{P\left(JI\cdot P\left(B\mid JI\right)\right)}{P\left(B\right)}$

or,

$P\left(JI\mid B\right)=\frac{P\left(JI\right)\cdot P\left(B\mid JI\right)}{P\left(JI\right)\cdot P\left(B\mid JI\right)+P\left(J\text{II}\right)\cdot P\left(B\mid J\text{II}\right)}$

The last statement is Bayes' Formula for the case where the sample space is divided into two partitions. The following is the generalization of this formula for n partitions.

We begin with the following example.

A department store buys 50% of its appliances from Manufacturer A, 30% from Manufacturer B, and 20% from Manufacturer C. It is estimated that 6% of Manufacturer A's appliances, 5% of Manufacturer B's appliances, and 4% of Manufacturer C's appliances need repair before the warranty expires. An appliance is chosen at random. If the appliance chosen needed repair before the warranty expired, what is the probability that the appliance was manufactured by Manufacturer A? Manufacturer B? Manufacturer C?

Let events $A$ , $B$ and $C$ be the events that the appliance is manufactured by Manufacturer A, Manufacturer B, and Manufacturer C, respectively. Further, suppose that the event $R$ denotes that the appliance needs repair before the warranty expires.

We need to find $P\left(A\mid R\right)$ , $P\left(B\mid R\right)$ and $P\left(C\mid R\right)$ .

We will do this problem both by using a tree diagram and by using Bayes' formula.

We draw a tree diagram.

The probability $P\left(A\mid R\right)$ , for example, is a fraction whose denominator is the sum of all probabilities of all branches of the tree that result in an appliance that needs repair before the warranty expires, and the numerator is the branch that is associated with Manufacturer A. $P\left(B\mid R\right)$ and $P\left(C\mid R\right)$ are found in the same way. We list both as follows:

$P\left(A\mid R\right)=\frac{\text{.}\text{030}}{\left(\text{.}\text{030}\right)+\left(\text{.}\text{015}\right)+\left(\text{.}\text{008}\right)}=\frac{\text{.}\text{030}}{\text{.}\text{053}}=\text{.}\text{566}$

$P\left(B\mid R\right)=\frac{\text{.}\text{015}}{\text{.}\text{053}}=\text{.}\text{283}$ and $P\left(C\mid R\right)=\frac{\text{.}\text{008}}{\text{.}\text{053}}=\text{.}\text{151}$ .

Alternatively, using Bayes' formula,

$\begin{array}{c}P\left(A\mid R\right)=\frac{P\left(A\right)P\left(R\mid A\right)}{P\left(A\right)P\left(R\mid A\right)+P\left(B\right)P\left(R\mid B\right)+P\left(C\right)P\left(R\mid C\right)}\\ =\frac{\text{.}\text{030}}{\left(\text{.}\text{030}\right)+\left(\text{.}\text{015}\right)+\left(\text{.}\text{008}\right)}=\frac{\text{.}\text{030}}{\text{.}\text{053}}=\text{.}\text{566}\end{array}$

$P\left(B\mid R\right)$ and $P\left(C\mid R\right)$ can be determined in the same manner.

Got questions? Get instant answers now!

Got questions? Get instant answers now!