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horrorfan [7]
3 years ago
5

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

en the couple will have if, on Mars, males are:
(a) Half as likely as females.
(b) Just as likely as females.
(c) Twice as likely as females.
Mathematics
1 answer:
Ne4ueva [31]3 years ago
5 0

Answer:

a) 6

b) 4

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,

a1= number of trials performed until 2 males are born in experiment 1

a2= number of trials performed until 2 males are born in experiment 2

and so on.

If a1 + a2 + … + an = N

we would expect Np females.  

Since the experiment was performed n times, there 2n males (recall that the experiment stops when 2 males are born).

So we would expect 2n = N(1-p), or

N/n = 2/(1-p)

But N/n is the average number of trials per experiment, that is, the expectation.

<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

The expected number of trials would be

2/(1/2) = 4

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

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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Answer:

See explanation and figures below.

Step-by-step explanation:

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For this case the original function is y = x^2 with a vertex (0,0) for the new function we have the following general expression h(x)= (x-h)^2 + k on this case we see that h = -2 and k =-4  so then the vertex for the new function would be (-2,4), and the x intercepts are (-4,0) and (0,0) as we can see on the figure attached 1.

b. ℎ(x) = −(x − 4)^2 + 2

For this case the original function is y = x^2 with a vertex (0,0) for the new function we have the following general expression h(x)= (x-h)^2 + k on this case we see that h = 4 and k =2  so then the vertex for the new function would be (4,2), and the x intercepts are (2.586,0) and (5.414,0) as we can see on the figure attached 2.

On this case since we have a negative on the quadratic term we see that the parabola would open downwards.  

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e. h(x) = 3x^2 + 6x

For this case after the first transformation 3x^2 we have a compression on the x axis, the +6x would cause that the function move to the left and down and an expansion on the x axis. The new vertex would be (-1,-3) and the x intercepts would be:

3x(x+2) = 0

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And we can see the plot on figure 5 attached.

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Step-by-step explanation:

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3 years ago
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