Which expression has a value of 35 when p = 7

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

Hey, I'm bo.red anyone wanna pl.ay Fort.nite with me? :,)

Karo-lina-s [1.5K]

Answer:

hi

Step-by-step explanation:

yes how you doing tonight

6 0

2 years ago

Read 2 more answers

Please teach me on how to answer this question. (addmath question)

arsen [322]

We know that $\overline{x}=8$ so:

$\overline{x}=8\\\\\\\dfrac{\sum\limits_{k=1}^7\,x_k}{7}=8\qquad|\cdot7\\\\\\ \boxed{\sum\limits_{k=1}^7\,x_k=56}$

We want to calculate:

$\sum\limits_{k=1}^7\,\big(2x_k-3\big)^2=\sum\limits_{k=1}^7\,\big(4x_k^2-12x_k+9\big)=\\\\\\=\sum\limits_{k=1}^7\,4x_k^2-\sum\limits_{k=1}^7\,\big12x_k+\sum\limits_{k=1}^7\,9=4\sum\limits_{k=1}^7\,x_k^2-12\sum\limits_{k=1}^7\,x_k+\sum\limits_{k=1}^7\,9=\\\\\\=4\cdot672-12\cdot56+7\cdot9=2688-672+63=\boxed{2079}$