Question

Mark is deciding which route to take to work. His choices are I = the Interstate and F = Fifth Street.

•$P\left(I\right)=0.44andP\left(F\right)=0.56$

• role="math" localid="1652098516712" $P(I\cap F)=0$because Mark will take only one route to work.

What is the probability of $P(IORF)$?

Short answer

$P(IorF)=1$

Step-by-step solution

Basic Concept

Probability is the likelihood of an event.

As per probability rule,

$P(AUB)=P\left(A\right)+P\left(B\right)-P(A\cap B)$

Explanation 

It is given that Mark takes Interstate or fifth street route to work, with$Pr\left(I\right)\mathrm{and}Pr\left(F\right)=0.44\mathrm{and}0.56$ respectively.

As he can take only one route & both routes can never be taken together, so these events are mutually exclusive, represented by disjoint sets whose intersection i.e.$P(I\&F)=P(I\cap F)=0$

$So,P(IorF)=P(IUF)=P\left(I\right)+P\left(F\right)-P(I\cap F)$

$P(IorF)=0.44+0.56=1$

This shows - it is certain that Mark will take either route to work