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daser333 [38]
3 years ago
9

A presidential candidate plans to begin her campaign by visiting the capitals in 4 of 42 states. What is the probability that sh

e selects the route of four specific​ capitals? Is it practical to list all of the different possible routes in order to select the one that is​ best?
P(she selects the route of four specific ​capitals): _____.
Is it practical to list all of the different possible routes in order to select the one that is​ best?
A.
​Yes, it is practical to list all of the different possible routes because the number of possible permutations is very small.
B.
​Yes, it is practical to list all of the different possible routes because the number of possible permutations is very large.
C.
No, it is not practical to list all of the different possible routes because the number of possible permutations is very small.
D.
No, it is not practical to list all of the different possible routes because the number of possible permutations is very large.
Mathematics
1 answer:
marissa [1.9K]3 years ago
7 0

Answer:

P (She selects the route of four specific capitals) = \frac{1}{2686320}=(3.7226)10^{-7}

D. No,it is not practical to list all of the different possible routes because the number of possible permutations is very large.

Step-by-step explanation:

Let's start assuming that each route is equally likely to be chosen.

Assuming this, we can calculate P(A) where the event A is ''She selects the route of four specific capitals'' doing the following :

P(A) = Favourable cases in which the route of four specific capitals is selected / Total number of ways in 4 of 42 states

The favourable cases in which the route of four specific capitals is selected is equal to 1 .

For the denominator we need the permutation number of 4 in 42.

The permutation number is defined as :

nPr=\frac{n!}{(n-r)!}

42P4=\frac{42!}{(42-4)!}=\frac{42!}{38!}=2686320

The probability of event A is : \frac{1}{2686320}=(3.7226)10^{-7}

Finally for the other question : The option D is the correct because the number of possible permutations is 2686320 and is very large to be listed.

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