Help me please *an image is attached*

You play the following game against your friend. You have 2 urns and 4 balls One of the balls is black and the other 3 are white

Rom4ik [11]

Answer:

Part a: The case in such a way that the chances are minimized so the case is where all the four balls are in 1 of the urns the probability of her winning is least as 0.125.

Part b: The case in such a way that the chances are maximized so the case where the black ball is in one of the urns and the remaining 3 white balls in the second urn than, the probability of her winning is maximum as 0.5.

Part c: The minimum and maximum probabilities of winning for n number of balls are such that

when all the n balls are placed in one of the urns the probability of the winning will be least as 1/2n

when the black ball is placed in one of the urns and the n-1 white balls are placed in the second urn the probability is maximum, as 0.5

Step-by-step explanation:

Let us suppose there are two urns A and A'. The event of selecting a urn is given as A thus the probability of this is given as

P(A)=P(A')=0.5

Now the probability of finding the black ball is given as

P(B)=P(B∩A)+P(P(B∩A')

P(B)=(P(B|A)P(A))+(P(B|A')P(A'))

Now there can be four cases as follows

Case 1: When all the four balls are in urn A and no ball is in urn A'

so

P(B|A)=0.25 and P(B|A')=0 So the probability of black ball is given as

P(B)=(0.25*0.5)+(0*0.5)

P(B)=0.125;

Case 2: When the black ball is in urn A and 3 white balls are in urn A'

so

P(B|A)=1.0 and P(B|A')=0 So the probability of black ball is given as

P(B)=(1*0.5)+(0*0.5)

P(B)=0.5;

Case 3: When there is 1 black ball and 1 white ball in urn A and 2 white balls are in urn A'

so

P(B|A)=0.5 and P(B|A')=0 So the probability of black ball is given as

P(B)=(0.5*0.5)+(0*0.5)

P(B)=0.25;

Case 4: When there is 1 black ball and 2 white balls in urn A and 1 white ball are in urn A'

so

P(B|A)=0.33 and P(B|A')=0 So the probability of black ball is given as

P(B)=(0.33*0.5)+(0*0.5)

P(B)=0.165;

Part a:

As it says the case in such a way that the chances are minimized so the case is case 1 where all the four balls are in 1 of the urns the probability of her winning is least as 0.125.

Part b:

As it says the case in such a way that the chances are maximized so the case is case 2 where the black ball is in one of the urns and the remaining 3 white balls in the second urn than, the probability of her winning is maximum as 0.5.

Part c:

The minimum and maximum probabilities of winning for n number of balls are such that

when all the n balls are placed in one of the urns the probability of the winning will be least given as

P(B|A)=1/n and P(B|A')=0 So the probability of black ball is given as

P(B)=(1/n*1/2)+(0*0.5)

P(B)=1/2n;

when the black ball is placed in one of the urns and the n-1 white balls are placed in the second urn the probability is maximum, equal to calculated above and is given as

P(B|A)=1/1 and P(B|A')=0 So the probability of black ball is given as

P(B)=(1/1*1/2)+(0*0.5)

P(B)=0.5;

5 0

3 years ago

Find the value of x 50 x

finlep [7]

The value of X is 80

6 0

2 years ago

What is the following product? 3^sqr4 * sqr3

Nina [5.8K]

Answer:

9sqrt3

Step-by-step explanation:

This is an order of operations question.

First, simply the exponent, which gives you 3^2*sqrt3

Next, 3^2=9, giving the final answer of 9sqrt3.

3 0

3 years ago

Find the horizontal and vertical asymptotes of f(x). f(x) equals = StartFraction 6 x Over x plus 2 EndFraction 6x x+2 Find the

miss Akunina [59]

Answer:

The horizontal asymptote can be described by the line y = 6

The vertical asymptote can be described by the line x = -2

Step-by-step explanation:

* Lets the meaning of vertical and horizontal asymptotes

- Vertical asymptotes are vertical lines which correspond to the zeroes

of the denominator of a rational function

- A horizontal asymptote is a y-value on a graph which a function

approaches but does not actually reach

- If the degree of the numerator is less than the degree of the

denominator, then there is a horizontal asymptote at y = 0

- If the degree of the numerator is greater than the degree of the

denominator, then there is no horizontal asymptote

- If the degree of the numerator is equal the degree of the denominator,

then there is a horizontal asymptote at y = leading coefficient of the

numerator ÷ leading coefficient of the denominator

* Lets solve the problem

∵ $f(x)=\frac{6x}{x+2}$

∵ The numerator is 6x

∵ The denominator is x + 2

∴ The numerator and the denominator have same degree

∵ The leading coefficient of the numerator is 6

∵ The leading coefficient of the denominator is 1

∴ There is a horizontal asymptote at y = 6/1

∴ The horizontal asymptote can be described by the line y = 6

- Put the denominator equal zero to find its zeroes

∵ The denominator is x + 2

∴ x + 2 = 0

- Subtract 2 from both sides

∴ x = -2

∴ The vertical asymptote can be described by the line x = -2

6 0

3 years ago

Read 2 more answers

Spiral Review At a restaurant, Brianna and Naomi each order an appetizer for $6.50 and an entree for $9.75. They decide to leave

Licemer1 [7]

Answer: 4.55

Step-by-step explanation:

6.50x2 = 13 + 9.75 = 20% = 4.55

hoped this helped lol

4 0

3 years ago