Q.3.11 - A First Course in Probability

Question

A type $C$ battery is in working condition with probability .$7$, whereas a type $D$ battery is in working condition with probability .$4$. A battery is randomly chosen from a bin consisting of $8$ type C and $6$ type $D$ batteries. 

(a) What is the probability that the battery works? 

(b) Given that the battery does not work, what is the conditional probability that it was a type $C$ battery? 

Step-by-Step Solution

Answer

a). The probability that the battery works is $\frac{4}{7}$.

b). The conditional probability that it was a type $C$ battery is $\frac{6}{15}$.

Step 1: Given Information (Part a)

$C-C$ battery was chosen

$D-D$ battery was chosen

$W$ - the battery works

Given probabilities:

$P(W\mid C)=0.7$

$P(W\mid D)=0.4$

$P\left(C\right)=\frac{8}{8+6}=\frac{4}{7}$

$P\left(D\right)=\frac{6}{8+6}=\frac{3}{7}$

Step 2: Explanation (Part a)

The formula of total probability can be applied here (because $C \cap D=\varnothing$, $P(C \cap D)=1 )$:

$P\left(W\right)=P(W\mid C)P\left(C\right)+P(W\mid D)P\left(D\right)$

Substitution of familiar probabilities:

$P\left(W\right)=0.7·\frac{4}{7}+0.4·\frac{3}{7}=\frac{4}{7}$

Step 3: Final Answer (Part a)

The probability that the battery works is $\frac{4}{7}$. 

Step 4: Given Information (Part b)

$P(W\mid C)=0.7$

$P(W\mid D)=0.4$

$P\left(C\right)=\frac{8}{8+6}=\frac{4}{7}$

$P\left(D\right)=\frac{6}{8+6}=\frac{3}{7}$

Step 5: Explanation (Part b)

The definition of conditional probability gives:

$P\left(C\mid W^c\right)=\frac{P\left(C\cap W^c\right)}{P\left(W^c\right)} \text{ and } P\left(W^c\mid C\right)·P\left(C\right)=P\left(W^c\cap C\right)$

Which can be fused into:

$P\left(C\mid W^c\right)=\frac{P\left(W^c\mid C\right)·P\left(C\right)}{P\left(W^c\right)}$

Now formula for complement (with both probability and conditional probability):

$P\left(W^c\right)=1-P\left(W\right)=1-\frac{4}{7}=\frac{3}{7}$$P\left(W^c\mid C\right)=1-P(W\mid C)=1-0.7=0.3$

Substitution of this:

$P\left(C\mid W^c\right)=\frac{0.3·\frac{4}{7}}{\frac{3}{7}}=\frac{6}{15}$

Step 6: Final Answer (Part b)

The conditional probability that it was a type $C$ battery is $\frac{6}{15}$.