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3 years ago

15

In a lottery game, a player picks six numbers from 1 to 22. If the player matches all six numbers, they win 20,000 dollars. Othe

rwise, they lose $1.
What is the expected value of this game?

1 answer:

Angelina_Jolie [31]3 years ago

8 0

Answer:

Total possible number of outcomes = C(24,6) [24 choose 6]

=24!/(6!18!)

= 134596

Out of which there is only one winning combination.

Therefore we conclude:

P(win 20000)=1/134596

P(lose 1)=134595/134596

and hence the expected value is:

20000*(1/134596)+(-1)*(134595/134596)

=-114595/134596

=-0.8514 (rounded to four places after decimal)

Step-by-step explanation:

Hope this helped!

You might be interested in

A company makes two products from steel; one requires 2 tons of steel and the other requires 3 tons. There are 100 tons of steel

LenaWriter [7]

Answer:

true

Step-by-step explanation:

first company requires 2 tons of steel

Second company requires 3 tons of steel.

Total amount of steel is 100 tons

So, te total amount of steel required by both the companies should be equal to less than 100.

S, the constraint is

$2x_{1}+3x_{2}$

Thus, the given relation is true.

6 0

4 years ago

Chose parameters h and k such that the system has a) a unique solution, b) many solutions, and c) no solution. X1 + 3x2 = 4 2x1

AysviL [449]

Answer:

a) The system has a unique solution for $k\neq 6$ and any value of $h$ , and we say the system is consisted

b) The system has infinite solutions for $k=6$ and $h=8$

c) The system has no solution for $k=6$ and $h\neq 8$

Step-by-step explanation:

Since we need to base the solutions of the system on one of the independent terms ( $h$ ), the determinant method is not suitable and therefore we use the Gauss elimination method.

The first step is to write our system in the augmented matrix form:

$\left[\begin{array}{cc|c}1&3&4\\2&k&h\end{array}\right]$

The we can use the transformation $r_0\rightarrow r_0 -2r_1$ , obtaining:

$\left[\begin{array}{cc|c}1&3&4\\0&k-6&h-8\end{array}\right]$ .

Now we can start the analysis:

If $k\neq 6$ then, the system has a unique solution for any value of $k$ , meaning that the last row will transform back to the equation as:

$(k-6)x_2=h-8\\x_2=h-8/(k-6)$

from where we can see that only in the case of $k=6$ the value of $x_2$ can not be determined.

if $k=6$ and $h=8$ the system has infinite solutions: this is very simple to see by substituting these values in the equation resulting from the last row:

$(k-6)x_2=h-8\\0=0$ which means that the second equation is a linear combination of the first one. Therefore, we can solve the first equation to get $x_1$ as a function of $x_2$ o viceversa. Thus, $x_2$ ( $x_1$ ) is called a parameter since there are no constraints on what values they can take on.

if $k=6$ and $h\neq 8$ the system has no solution. Again by substituting in the equation resulting from the last row:

$(k-6)x_2=h-8\\0=h-8$ which is false for all values of $h\neq 8$ and since we have something that is not possible $(0\neq h-8,\ \forall \ h\neq 8)$ the system has no solution

6 0

4 years ago

Find the Correlation of the following two variables X: 2, 3, 5, 6 Y: 1, 2, 4, 5

Karolina [17]

Answer:

The correlation of X and Y is 1.006

Step-by-step explanation:

Given

X: 2, 3, 5, 6

Y: 1, 2, 4, 5

n = 4

Required

Determine the correlation of x and y

Start by calculating the mean of x and y

<em>For x</em>

$M_x = \frac{\sum x}{n}$

$M_x = \frac{2 + 3+5+6}{4}$

$M_x = \frac{16}{4}$

$M_x = 4$

<em>For y</em>

$M_y = \frac{\sum y}{n}$

$M_y = \frac{1+2+4+5}{4}$

$M_y = \frac{12}{4}$

$M_y = 3$

Next, we determine the standard deviation of both

$S = \sqrt{\frac{\sum (x - Mean)^2}{n - 1}}$

For x

$S_x = \sqrt{\frac{\sum (x_i - Mx)^2}{n -1}}$

$S_x = \sqrt{\frac{(2-4)^2 + (3-4)^2 + (5-4)^2 + (6-4)^2}{4 - 1}}$

$S_x = \sqrt{\frac{-2^2 + (-1^2) + 1^2 + 2^2}{3}}$

$S_x = \sqrt{\frac{4 + 1 + 1 + 4}{3}}$

$S_x = \sqrt{\frac{10}{3}}$

$S_x = \sqrt{3.33}$

$S_x = 1.82$

For y

$S_y = \sqrt{\frac{\sum (y_i - My)^2}{n - 1}}$

$S_y = \sqrt{\frac{(1-3)^2 + (2-3)^2 + (4-3)^2 + (5-3)^2}{4 - 1}}$

$S_y = \sqrt{\frac{-2^2 + (-1^2) + 1^2 + 2^2}{3}}$

$S_y = \sqrt{\frac{4 + 1 + 1 + 4}{3}}$

$S_y = \sqrt{\frac{10}{3}}$

$S_y = \sqrt{3.33}$

$S_y = 1.82$

Find the N pairs as $(x-M_x)*(y-M_y)$

$(2 - 4)(1 - 3) = (-2)(-2) = 4$

$(3 - 4)(2 - 3) = (-1)(-1) = 1$

$(5 - 4)(4 - 3) = (1)(1) = 1$

$(6-4)(5-3) = (2)(2) = 4$

Add up these results;

$N = 4 + 1 + 1 + 4$

$N = 10$

Next; Evaluate the following

$\frac{N}{S_x * S_y} * \frac{1}{n-1}$

$\frac{10}{1.82* 1.82} * \frac{1}{4-1}$

$\frac{10}{3.3124} * \frac{1}{3}$

$\frac{10}{9.9372}$

$1.006$

<em>Hence, The correlation of X and Y is 1.006</em>

4 0

4 years ago

zalisa [80]

Answer:

true;)

Step-by-step explanation:

its right

4 0

3 years ago

Read 2 more answers

How many sides does a rhombicosidodecahedron have

Sveta_85 [38]

Answer:

It has 120 sides but mark the other guy brainliest

Step-by-step explanation:

8 0

3 years ago