Answer
a)0.02
b)0.0004
c)0.000008.
d)Very big improvement.
Work Step by Step
a) Here, 2%$=\frac{2}{100}=0.02.$
b) Events are dependent, if the outcome of one effects the outcome of the other. Here the events are independent, because the failure of one doesn't affect the other, hence here $P(B|A)=P(B)$. $P(A \cap B)=P(A)\cdot P(B|A)$. Hence $P(A \cap B)=\frac{2}{100}\cdot\frac{2}{100}=0.0004.$
c) Events are dependent, if the outcome of one effects the outcome of the other. Here the events are independent, because the failure of one doesn't affect the other, hence here $P(B|A)=P(B)$. $P(A \cap B \cap C)=P(A)\cdot P(B|A)\cdot P(C|(A \cap B))$. Hence $P(A \cap B \cap C)=\frac{2}{100}\cdot\frac{2}{100}\cdot\frac{2}{100}=0.000008.$
d) Changing from one drive to 3 reduces the probability of failure from 0.02 to 0.000008, which is a very big improvement.
