Question

A survey of 1,108 employees at a software company finds that 621 employees take a bus to work and 445 employees take a train to work. Some employees take a bus and a train, and 321 employees only take a train. To the nearest percent, find the probability that a randomly chosen employee takes a bus or a train to work. Explain.

Answer 1

We know 445 employees take the train, and that 321 of these exclusively take the train. So 445 - 321 = 124 take both the train and bus.

Now, if $B$ is the set of employees that take the bus and $T$ the set of employees that take the train, then

$n(B\cup T)=n(B)+n(T)-n(B\cap T)=621+445-124=942$

where $n(A)$ is the number of employees belonging to a general set $A$.

So the probability that an employee takes either the bus or train is

$\dfrac{n(B\cup T)}{1108}=\dfrac{942}{1108}\approx85\%$