Question

4 republicans, 8 democrats, and 5 independents are randomly rated in a row. What is the probability that

1. the first 5 seats are taken by the democrats?

2. none of the first 5 seats is taken by democrats?

3. the final 3 seats are taken by a republican, and an independent

4.all republicans are seated together

Answer 1

nPr = n! / (n-r)!

Total Number of people = 4 + 8 + 5 = 17

Probability = Favorable Outcomes / Total Outcomes

Total Outcomes = 17!

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(a) The first 5 seats can be seated in 8P5 ways

The remaining 12 seats can be arranged in 12! ways.

Therefore Total Favorable Outcomes = 8P5 * 12!

The Required Probability = (8P5 * 12!) / 17! = 0.009

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(b) The probability that none of the 5 seats are taken by democrats = 1 - P(all seats taken by democrats) = 1 - 0.009 = 0.991

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(c) Republicans can be chosen for 2 of the 3 seats in 4C2 ways and and Independent can be chosen for 1 seat in 5C1 ways

These 3 seats can then be arranged in 3! ways.

The remaining 14 seats can be arranged in 14! ways

Therefore Total Favorable Outcomes = 4C2 * 4C1 * 3! * 14!

The Required Probability = (4C2 * 4C1 * 3! * 14!) / 17! = 0.0353

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(d) Let us take all the 4 republicans as 1 unit. Therefore we are now arranging 1 unit of the republicans + 8 democrats + 5 independents = 14 people in 14! ways.

The republican themselves can be arranged in 4! ways.

Therefore Total Favorable Outcomes = 14! * 3!

The Required Probability = (14! * 3!) / 17! = 0.0015

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