Question

The probability that a new car battery functions for more than $10,000$miles is $8$, the probability that it functions for more than $20,000$miles is$4$, and the probability that it functions for more than $30,000$miles is $1$. If a new car battery is still working after$10,000$ miles, what is the probability that
$\left(a\right)$its total life will exceed$20,000$miles?
$\left(b\right)$its additional life will exceed $20,000$miles?

Short answer

(a) The probability of total life will exceed $20000$miles is $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right..$

(b) The probability of additional life will exceed $20000$miles is $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$8$}\right..$

Step-by-step solution

Given values

The battery works after$10,000$miles - $A_1.$

The battery works after $20,000$miles -$A_2.$

The battery works after $30,000$miles - $A_3.$

The probabilities are

$\begin{array}{c}P\left(A_1\right)=0.8.\\ P\left(A_2\right)=0.4.\\ P\left(A_3\right)=0.1.\end{array}$

The nature yields events are

$\begin{array}{c}\Rightarrow A_3\subseteq A_2\subseteq A_1.\\ \Rightarrow A_1\cap A_2=A_2.\\ \Rightarrow A_1\cap A_3=A_3.\end{array}$

For certain battery, the worked after $x$and$y$.if$x\le y.$

Find the probability that the total life will exceed 20000 miles (part a)

By conditional probability,

$$\begin{gathered} P\left(A_2\mid A_1\right)=\frac{P\left(A_2\cap A_1\right)}{P\left(A_1\right)} \\ P\left(A_2\mid A_1\right)=\frac{P\left(A_2\right)}{P\left(A_1\right)} \\ P\left(A_2\mid A_1\right)=\frac{0.4}{0.8} \\ P\left(A_2\mid A_1\right)=\frac{1}{2}. \end{gathered}$$

Find the probability that the additional life will exceed 20000 miles  (part b)

By conditional probability,

$$\begin{gathered} P\left(A_3\mid A_1\right)=\frac{P\left(A_3\cap A_1\right)}{P\left(A_1\right)} \\ P\left(A_3\mid A_1\right)=\frac{P\left(A_3\right)}{P\left(A_1\right)} \\ P\left(A_3\mid A_1\right)=\frac{0.1}{0.8} \\ P\left(A_3\mid A_1\right)=\frac{1}{8}. \end{gathered}$$