Question

 Let A and B be two events defined on a sample space S. If Pr(A)=0.8; Pr(A| B )=0.85 and Pr(A| B^ c )=0.75; determine the probability that neither of the two events occur.

Answer 1

The probability that neither of these events occur is: $P(\bar{A}\cap\bar{B})$, We assume that $B^c=\bar{B}.$ By definition, we have $P(A|\bar{B})=\frac{P(A\cap \bar{B})}{P(\bar{B})}$. $P(A|{B})=\frac{P(A\cap {B})}{P({B})}$. The following formula holds: $P(\bar{B})=P({A}\cap \bar{B})+P({\bar{A}}\cap {\bar{B}})$. From the latter we get: $P(\bar{B})-P({A}\cap \bar{B})=P({\bar{A}}\cap {\bar{B}})$. $P({A}\cap \bar{B})=P(A|\bar{B})P(\bar{B})$. Thus, we receive that: $P({\bar{A}}\cap {\bar{B}})=P(\bar{B})(1-P(A|\bar{B}))=(1-P({B})(1-P(A|\bar{B}))$.$P(A)=P({A}\cap \bar{B})+P({{A}}\cap {{B}})$. From formulae for $P(A|\bar{B})$ and $P(A|{B})$ we receive: $P(A)=P(A|\bar{B})(1-P({B}))+P(A|{B})P(B)$. From the latter we receive:$P(B)=\frac{(P(A)-P(A|\bar{B}))}{P(A|B)-P(A|\bar{B})}$. Thus, we have: $P(B)=\frac{0.8-0.75}{0.85-0.75}=0.5$. Substitute the latter into expression for $P({\bar{A}}\cap {\bar{B}})$: $P({\bar{A}}\cap {\bar{B}})=0.5\cdot0.25=0.125.$ Thus, the probability, that neither of events occur is: $0.125.$