Q. 3.2

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
$(a)$ its total life will exceed $20, 000$ miles?
$(b)$ its additional life will exceed $20, 000$ miles?

Step-by-Step Solution

Verified

Answer

(a) The probability of total life will exceed $20000$ miles is $\frac{1}{2} .$

(b) The probability of additional life will exceed $20000$ miles is $\frac{1}{8} .$

1Step1: 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

$\begin{matrix} P (A_{1}) = 0.8 . \\ P (A_{2}) = 0.4 . \\ P (A_{3}) = 0.1 . \end{matrix}$

The nature yields events are

$\begin{matrix} \Rightarrow A_{3} \subseteq A_{2} \subseteq A_{1} . \\ \Rightarrow A_{1} \cap A_{2} = A_{2} . \\ \Rightarrow A_{1} \cap A_{3} = A_{3} . \end{matrix}$

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

2Step2: Find the probability that the total life will exceed 20000 miles (part a)

By conditional probability,

$P (A_{2} ∣ A_{1}) = \frac{P (A_{2} \cap A_{1})}{P (A_{1})} P (A_{2} ∣ A_{1}) = \frac{P (A_{2})}{P (A_{1})} P (A_{2} ∣ A_{1}) = \frac{0.4}{0.8} P (A_{2} ∣ A_{1}) = \frac{1}{2} .$

3Step3: Find the probability that the additional life will exceed 20000 miles (part b)

By conditional probability,

$P (A_{3} ∣ A_{1}) = \frac{P (A_{3} \cap A_{1})}{P (A_{1})} P (A_{3} ∣ A_{1}) = \frac{P (A_{3})}{P (A_{1})} P (A_{3} ∣ A_{1}) = \frac{0.1}{0.8} P (A_{3} ∣ A_{1}) = \frac{1}{8} .$

Q. 3.31

Q. 3.3

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