Math Challenges

Solution:
\( \displaylines{\text{Let }D\text{ represent the event that a drive is defective, while }S\text{ represents a satisfactory drive.}\\ A\text{ is the event that brand A was chosen, while }B\text{ is the event that brand B was chosen.}\\ \text{We are asked to compute }P\left(A\left|D\right.\right).\text{ This has a formula, }P\left(A\left|D\right.\right)=\frac{P\left(A\cap D\right)}{P\left(D\right)}.\\ \text{Bayes Theorem also makes use of the fact that }P\left(A\cap D\right)=P\left(A\right)P\left(D\left|A\right.\right)\text{, giving us the formula}\\ P\left(A\left|D\right.\right)=\frac{P\left(A\right)P\left(D\left|A\right.\right)}{P\left(A\right)P\left(D\left|A\right.\right)+P\left(B\right)P\left(D\left|B\right.\right)}\\ \text{We were given the following data:}\\ P\left(A\right)=0.30,P\left(B\right)=0.70\\ P\left(D\left|A\right.\right)=0.55,P\left(D\left|B\right.\right)=0.05\\ \text{Thus}\\ P\left(A\left|D\right.\right)=\frac{0.30\cdot0.55}{0.30\cdot0.55+0.7\cdot0.05}=\frac{165}{200}=\frac{33}{40}} \)