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Gnoma [55]
3 years ago
11

A stable roommate problem with 4 students a, b, c, d is defined as follows. Each student ranks the other three in strict order o

f preference. A matching is defined as the separation of the students into two disjoint pairs. A matching is stable if no two separated students prefer each other to their current roommates. Does a stable matching always exist
Mathematics
1 answer:
Fantom [35]3 years ago
4 0

Answer:

Each student ranks the other three in strict order of preference.

Step-by-step explanation:

When  4 different students from different background and personality lives together, there seems to be a room for the problems and stability between them. When the problem of stability arises, it would be as a result of the other 3 students ranking each other in strict order of preference regarding to their relationship and association.

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A Martian couple has children until they have 2 males (sexes of children are independent). Compute the expected number of childr
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a) 6

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c) 3

Step-by-step explanation:

Let p be the probability of having a female Martian, and of course, 1-p the probability of having a male Martian.

To compute the expected total number of trials before 2 males are born, imagine an experiment simulating the fact that 2 males are born is performed n times.

Let ak be the number of trials performed until 2 males are born in experiment k. That is,

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and so on.

If a1 + a2 + … + an = N

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<em>We have then that the expected number of trials before 2 males are born is 2/(1-p) where p is the probability of having a female. </em>

a)

Here we have the probability of having a male is half as likely as females. So

1-p = p/2 hence p=2/3

The expected number of trials would be

2/(1-2/3) = 2/(1/3) =6

This means <em>the couple would have 6 children</em>: 4 females (the first 4 trials) and 2 males (the last 2 trials).

b)

Here the probability of having a female = probability of having a male = 1/2

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c)

Here, 1-p = 2p so p=1/3

The expected number of trials would be

2/(1-1/3) = 2/(2/3) = 6/2 =3

This means<em> the couple would have 3 children</em>: 1 female (the first trial) and 2 males (the last 2 trials).

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