There are 10 students in a classroom, including the triplets Hugh, Stu, and Lou. If 3 of the 10 are randomly selected to give speeches, what is the probability that Hugh will be first, Stu will be second and Lou will be third?

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There are 10 students in a classroom, including the triplets Hugh, Stu, and Lou. If 3 of the 10 are randomly selected to give speeches, what is the probability that Hugh will be first, Stu will be second and Lou will be third?

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Answer 1 · 2017-07-17T03:37:59

$P (Hugh first, Stu second, Lou third) = \frac{1}{720} ≅ .0014$ $P("Hugh first, Stu second, Lou third")=1/720~=.0014$

Explanation:

This question is asking about permutations - randomly picking elements from a population where the order of the picks matters. The general formula is:

$P_{n, k} = \frac{n!}{(n - k)!}; n = population, k = picks$ $P_(n,k)=(n!)/((n-k)!); n="population", k="picks"$

We are interested in permutation involving having Hugh be first, Stu second, and Lou third. There is only 1 permutation where that can happen.

There are a total of $P_{10, 3}$ $P_(10,3)$ permutations that can be made:

$P_{10, 3} = \frac{10!}{(10 - 3)!} = \frac{10!}{7!} = \frac{10 \times 9 \times 8 \times 7!}{7!} = 10 \times 9 \times 8 = 720$ $P_(10,3)=(10!)/((10-3)!)=(10!)/(7!)=(10xx9xx8xx7!)/(7!)=10xx9xx8=720$

And so:

$P (Hugh first, Stu second, Lou third) = \frac{1}{720} ≅ .0014$ $P("Hugh first, Stu second, Lou third")=1/720~=.0014$

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There are 10 students in a classroom, including the triplets Hugh, Stu, and Lou. If 3 of the 10 are randomly selected to give speeches, what is the probability that Hugh will be first, Stu will be second and Lou will be third?

1 Answer

Explanation:

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