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2 years ago

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Mathematics

1 answer:

barxatty [35]2 years ago

8 0

Answer:

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Suppose we play the following game based on tosses of a fair coin. You pay me $10, and I agree to pay you $n 2 if heads comes up

Artyom0805 [142]

Answer:

In the long run, ou expect to lose $4 per game

Step-by-step explanation:

Suppose we play the following game based on tosses of a fair coin. You pay me $10, and I agree to pay you $n^2 if heads comes up first on the nth toss.

Assuming X be the toss on which the first head appears.

then the geometric distribution of X is:

X $\sim$ geom(p = 1/2)

the probability function P can be computed as:

$P (X = n) = p(1-p)^{n-1}$

where

n = 1,2,3 ...

If I agree to pay you $n^2 if heads comes up first on the nth toss.

this implies that , you need to be paid $\sum \limits ^{n}_{i=1} n^2 P(X=n)$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) = E(X^2)$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) =Var (X) + [E(X)]^2$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p}{p^2}+(\dfrac{1}{p})^2$ ∵ X $\sim$ geom(p = 1/2)

$\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p}{p^2}+\dfrac{1}{p^2}$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{1-p+1}{p^2}$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{2-p}{p^2}$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) = \dfrac{2-\dfrac{1}{2}}{(\dfrac{1}{2})^2}$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ \dfrac{4-1}{2} }{{\dfrac{1}{4}}}$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ \dfrac{3}{2} }{{\dfrac{1}{4}}}$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) =\dfrac{ 1.5}{{0.25}}$

$\sum \limits ^{n}_{i=1} n^2 P(X=n) =6$

Given that during the game play, You pay me $10 , the calculated expected loss = $10 - $6

= $4

∴

In the long run, you expect to lose $4 per game

3 0

3 years ago

What divide by what equals 0.4545......

Arturiano [62]

Answer:

5/11

Step-by-step explanation:

let x=0.4545...

100x=45.4545...

100x-x = 99x=45.4545... - 0.4545...=45

x=45/99=5/11

6 0

2 years ago

(+6) + (-11) + (+5)+(-7)

tensa zangetsu [6.8K]

Answer:

(+6) + (-11) + (+5)+(-7)

(6) + (-11) + (5) - (7)

6 - 11 + 5 - 7

(6 + 5) + (-11 - 7)

11 - 18

= -7

3 0

2 years ago

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Y= x^2 -5 solve for x

dexar [7]

$Y= x^2 -5$

We need to solve for x, we need to get x alone

$Y= x^2 -5$

Lets start by removing -5

Add 5 on both sides

$y + 5= x^2 -5 + 5$

$y + 5= x^2$

Now to isolate x , we need to remove the square from x

To remove square , take square root on both sides

$+-\sqrt{y+5} = \sqrt{x^2}$

square and square root will get cancelled

$+-\sqrt{y+5} = x$

So $x = +\sqrt{y+5}$ and $x = -\sqrt{y+5}$

7 0

2 years ago

Answer for -2x-41=5x-6

seraphim [82]

Answer:

x=-5

Step-by-step explanation:

-2x-41=5x-6

-35=7x

-5=x

7 0

2 years ago

Read 2 more answers