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?

$C-C$battery was chosen

$D-D$battery was chosen

$W$- the battery works

Given probabilities:

$P(W\mid C)=0.7$

$P(W\mid D)=0.4$

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

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

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:

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

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

$P(W\mid C)=0.7$

$P(W\mid D)=0.4$

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

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

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):

localid="1648004809650" $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:

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

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