## Precalculus (6th Edition) Blitzer

a) Probability of a flood two years in a row is $0.04$. b) Probability of a flood for three consecutive years is $0.008$. c) The probability of no flooding for four consecutive years is $0.4096$.
(a) We know that the probability of a flood in a given year in a region is, ${{\text{P}}_{\text{One year}}}\text{ = 0}\text{.2}$ And the probability of no flooding in a given year in a region is, \begin{align} & {{\text{P}}_{\text{No flooding}}}\text{ = 1}-{{\text{P}}_{\text{One year}}}\text{ } \\ & \text{= 1}-\text{0}\text{.2} \\ & \text{= 0}\text{.8} \end{align} Then, the probability of a flood two years in a row, \begin{align} & {{\text{P}}_{\text{Two years }}}={{\text{P}}_{\text{One year}}}\times \text{ }{{\text{P}}_{\text{One year}}} \\ & \text{= 0}\text{.2 }\times \text{ 0}\text{.2} \\ & \text{= 0}\text{.04} \end{align} (b) We know that the probability of a flood for three consecutive years is \begin{align} & {{\text{P}}_{\text{Three years }}}={{\text{P}}_{\text{One year}}}\times \text{ }{{\text{P}}_{\text{One year}}}\times \text{ }{{\text{P}}_{\text{One year}}} \\ & \text{= 0}\text{.2 }\times \text{ 0}\text{.2 }\times \text{ 0}\text{.2} \\ & \text{= 0}\text{.008} \end{align} (c) We know that the probability of no flooding for four consecutive years is \begin{align} & {{\text{P}}_{\text{Four years no flooding}}}\text{ = }{{\text{P}}_{\text{No flooding}}}\times {{\text{P}}_{\text{No flooding}}}\times {{\text{P}}_{\text{No flooding}}}\times {{\text{P}}_{\text{No flooding}}} \\ & =0.8\times 0.8\times 0.8\times 0.8 \\ & =0.4096 \end{align}