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A presidential candidate plans to begin her campaign by visiting the capitals in 3 of 42 states. What is the probability that she selects the route of three specific​ capitals?

Respuesta :

Answer as a fraction = 1/68880 which is exact

Answer in decimal form = 0.0000145 which is approximate

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Explanation:

Let's call these three cities A, B, and C

There is only one order ABC out of 42*41*40 = 68,880 different possible three city routes to pick from.

So the probability is 1/68880

If you want the answer in decimal form, then use your calculator to find

1/68880 = 0.0000145 approximately

Using the probability concept, it is found that there is a 0.000087 = 0.0087% probability that she selects the route of three specific​ capitals.

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  • A probability is the number of desired outcomes divided by the number of total outcomes.
  • The order in which the capitals are chosen is not important, which means that the combination formula is used to find the number of total outcomes.

Combination formula:

[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by:

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

In this problem:

  • 1 desired outcome.
  • The total number of outcomes is a combination of 3 from a set of 42, thus:

[tex]T = C_{42,3} = \frac{42!}{3!39!} = 11480[/tex]

The probability is:

[tex]p = \frac{1}{11480} = 0.000087[/tex]

0.000087 = 0.0087% probability that she selects the route of three specific​ capitals.

A similar problem is given at https://brainly.com/question/4037372

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