Suppose a relay team of 4 people will be chosen from the members of the Fleetfoo

Question

Suppose a relay team of 4 people will be chosen from the members of the Fleetfoot Running Club to compete in a county-wide race. There are 12 members of the club: 8 women and 4 men. (NOTE: a relay race is one in which each member of the team runs a portion of the race route alone. These parts are done sequentially and are called the “legs” of the race.) a) How many different possible teams of 4 could Fleetfoot send to the race? b) What is the probability that there is at least one man on the relay team? c) What is the probability that the relay team has more than two women? d) If we consider the order in which the relay team will run the legs of the race, how many ways can we set up the running order of the relay teams? e) Considering the order of the relay team, what is the probability that the first and last runner will be the same gender?

Explanation / Answer

a)

People are distinct. There are 12 of them and order matters, so

12P4 = 12!/(12-4)! = 11880 [ANSWER]

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b)
Let x = number of men (successes) in the sample.

Note that the probability of x successes out of n trials is          
          
P(x) = C(N-K, n-x) C(K, x) / C(N, n)          
          
where          
N = population size =    12      
K = number of successes in the population =    4      
n = sample size =    4      
x = number of successes in the sample =    0      
          
Thus,          
          
P(   0   ) =    0.141414141
Thus, P(at least 1) = 1 - P(0) = 0.858585859 [ANSWER]

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c)

Using a cumulative hypergeometric distribution table or technology, matching          
          
where          
N = population size =    12      
K = number of successes in the population =    8      
n = sample size =    4      
x = critical number of successes in the sample =    2      
          
Thus,          
P(at most   2   ) =    0.406060606 [ANSWER]

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d)

There are 4! = 24 ways [ANSWER]

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e)

There are (8P2 + 4P2) = 68 ways to choose the first and last so that they have the same gender.

There are 10P2 = 90 ways to choose anyone at the middle.

Hence, there are 90*68 = 6120 ways for this task.

Hence,

P = 6120/11880 = 0.515151515 [ANSWER]

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