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Lesechka [4]
3 years ago
12

Five men and 5 women are ranked according to their scores on an examination. Assume that no two scores are alike and all 10! pos

sible rankings are equally likely. Let X denote the highest ranking achieved by a woman. (For instance, X = 1 if the top-ranked person is female.) Find P { X = i } , i = 1 , 2 , 3 , … , 8 , 9 , 10 .
Mathematics
1 answer:
m_a_m_a [10]3 years ago
3 0

Answer:

P(X=1)=\frac{1}{2}\\P(X=2)=\frac{5}{18}\\P(X=3)=\frac{5}{36}\\P(X=4)=\frac{5}{84}\\P(X=5)=\frac{5}{252}\\P(X=6)=\frac{1}{252} \\P(X>6)=0

Step-by-step explanation:

Let's define the following event :

X : ''Highest ranking achieved by a woman''

There are five men and five women so the lowest ranking possible is 6, this means that all five men achieved a better ranking than all five women.

In terms of the variable :

X can assume the following values :

X=1\\X=2\\X=3\\X=4\\X=5\\X=6

⇒ P(X=7)=0\\P(X=8)=0\\P(X=9)=0\\P(X=10)=0

⇒ P(X>6)=0

Now let's calculate the probability function for X

P(X=1)=\frac{(5)(9!)}{(10!)}=0.5=\frac{1}{2}

We write on the denominator the total ways to arrange ten different persons in a line. This number is 10!

On the numerator we write the total cases in which the first person is a woman ⇒ (5).(9!)

Because they are five different women and when you pick one of then they are 9! ways to arrange the 9 remaining persons

Following this logic, we continue calculating the probability function :

P(X=2)=\frac{(5)(5)(8!)}{(10!)}=\frac{5}{18}

Again : Five different ways to choose between the five men in the first position.Then, 5 different ways to choose between the five women.And finally, 8! ways to arrange the remaining persons

P(X=3)=\frac{(5)(4)(5)(7!)}{(10!)}=\frac{5}{36}

Here we have : five different ways to choose between five men.Four different ways to choose between the four remaining men.Five different ways to choose between the five women for the third position and finally 7! ways to arrange the remaining persons.

P(X=4)=\frac{(5)(4)(3)(5)(6!)}{(10!)}=\frac{5}{84}

P(X=5)=\frac{(5)(4)(3)(2)(5)(5!)}{(10!)}=\frac{5}{252}

P(X=6)=\frac{(5)(4)(3)(2)(1)(5)(4!)}{(10!)}=\frac{1}{252}

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y = -3y -7 - 5  Substitute -3y-7 for x

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3 years ago
Choose the equivalent system of linear equations that will produce the same solution as the one given below.
Amanda [17]
<h2>Greetings!</h2>

Answer:

Option 4

Step-by-step explanation:

Firstly, we need to make the two x's or the two y numbers the same. Lets multiply the second equation by 2:

(2x + 3y = 5 ) * 2 =

4x + 6y = 10

Now, subtract the second equation from the first one:

(4x - y = -11) - (4x + 6y = 10)

4x - 4x = 0

-y - 6y = -7y

-11 - 10 = -21

-7y = -21

Divide both sides by -7:

y = 3

Plug 3 into the first equation as y:

4x - 3 = -11

4x = -11 + 3

4x = -8

x = -2

So x = -2 and y = 3. We need to find one of the options that equals this.


<h3>Option  1)</h3>

-4x - 9y = -21

-10y = -30

40x - 90y = -210

-90y = -270

40x - 0 = 40x

-90y - - 90y = -90 + 90 = 0y

-210 - -270 = -210 + 270 = 60

40x = 60

x = 1.5

We already know this isn't equivalent because the x value is not the x value of the given equation.


<h3>Option 2) </h3>

4x + 3y = 5

2y = -6

_______

8x + 6y = 10

6y = -18

8x - 0 = 8x

6y - 6y = 0

10 - -18 = 10 + 18 = 28

8x = 28

x = 3.5

So we know this isn't equivalent because again, the x value isn't the same as the given equation.


<h3>Option 3</h3>

7x - 3y = -11

9x = -6

-----------

63x - 27y = -99

63x = -42

----------------

63x - 63x = 0

-27y - 0 = -27y

-99 - - 42 = -99 + 42 = -57

-27y = -57

-9y = -19

9y = 19

y = 19/9

So we know this one isn't equivalent as the y value isn't the same as the given equation.


<h3>Option 4</h3>

12x - 3y = -33

14x = -28

x = -28/14 = -2

12(-2) - 3y = -33

-24 - 3y = -33

-3y = -33 + 24

-3y = -9

y = 9/3

y = 3

So x = -2 and y = -3 which is the same as the first equation.


<h2>Hope this helps!</h2>
3 0
3 years ago
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