Question

A presidential candidate plans to begin her campaign by visiting the capitals in 44 of 4747 states. What is the probability that she selects the route of fourfour specific​ capitals? Is it practical to list all of the different possible routes in order to select the one that is​ best?

Answer 1

we are given

A presidential candidate plans to begin her campaign by visiting the capitals in 4 of 47 states

so, the number of ways of selecting four specific capitals out of 47 states is

$=47C_4$

$=\frac{47!}{4!(47-4)!}$

$=178365$

now, we can find probability

so, the probability is

$=\frac{1}{178365}$..............Answer

Answer 2

Answer:

First, we need to find the total number of possible routes they can have. They want four possible routs about 47 states.

$C_{47}^{4} =\frac{47!}{(47-4)!} = \frac{47!}{43!}=\frac{47 \times 46 \times 45 \times 44 \times 43!}{43!}= 47 \times 46 \times 45 \times 44=4,280,760$

Therefore, there are 4,280,760 ways to select four routes.

The probability of having only one route is

$P=\frac{1}{4,280,760} \approx 0$

Additionally, it's not practical at all to list all different routes and selec the best one, because there are too many, if they listed all routes, the campaign will be over by then.