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In a bag there are a certain number of black balls and white balls. Th

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In a bag there are a certain number of black balls and white balls. Th  [#permalink]

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New post 07 Mar 2016, 06:46
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Question Stats:

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In a bag there are a certain number of black balls and white balls. The probability of picking up exactly 1 white ball when 2 balls are randomly drawn, is 1/2. Which of the following is the ratio of the number of black balls to white balls in the bag

A. 1:7
B. 2:1
C. 1:1
D. 4:1
E. 1:4
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Re: In a bag there are a certain number of black balls and white balls. Th  [#permalink]

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New post 07 Mar 2016, 07:58
since the probability of drawing a white ball out of two picks is 1/2.

the ratio of the white ball to black balls should be 1:1

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Re: In a bag there are a certain number of black balls and white balls. Th  [#permalink]

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New post 07 Mar 2016, 08:02
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vinoo7 wrote:
In a bag there are a certain number of black balls and white balls. The probability of picking up exactly 1 white ball when 2 balls are randomly drawn, is 1/2. Which of the following is the ratio of the number of black balls to white balls in the bag

A. 1:2
B. 2:1
C. 1:1
D. 4:1
E. 1:4


Hi vinoo7,

I do not think it is Sub -600 level by any chance, it is 700 level..

Also, the number of White and Black is interchangeable..
so ratio W/B= B/W
There can be various answers for this ..
One method could be..

Let the White be W and Black be B
so only one W in 2 balls would be \(\frac{W}{T}*\frac{B}{(T-1)}*2\)= 1/2..
So \(\frac{W}{T}*\frac{B}{(T-1)}= 1/4\)..

Lets first work on denominators


If you look at the denominator it is multiple of two consecutive numbers..
4 is not a multiple of 2 consecutive numbers....

the consecutive numbers will be..
3*4
4*5
5*6.. not possible as does not contain 4 as divisor
6*7.. not possible as does not contain 4 as divisor
7*8
8*9
9*10.. not possible as does not contain 4 as divisor

solutions now


1) If denominator = 3*4 numerator = 1*3
so \(\frac{W}{T}*\frac{B}{(T-1)}= 3/4*3\)
so WB=3 and W+B=4 so W and B can be 1 and 3..
so one ratio= 1:3 or 3:1

2)If denominator = 5*4 numerator = 1*5
so \(\frac{W}{T}*\frac{B}{(T-1)}= 5/4*5\)
so WB=5 and W+B=5 No values possible

3)If denominator = 7*8 numerator = 7*8/4=14
so \(\frac{W}{T}*\frac{B}{(T-1)}= 14/7*8\)
so WB=14 and W+B=8 No values possible

4) If denominator = 8*9 numerator = 8*9/4=18
so \(\frac{W}{T}*\frac{B}{(T-1)}= 18/8*9\)
so WB=18 and W+B=9 so W and B can be 6 and 3..
so another ratio= 6:3 or 3:6, which is 2:1 or 1:2..
so this is what we are looking for
ans can be A or B

Hope the method which we can adopt is clear

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Re: In a bag there are a certain number of black balls and white balls. Th  [#permalink]

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New post 04 Jul 2017, 00:33
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vinoo7 wrote:
In a bag there are a certain number of black balls and white balls. The probability of picking up exactly 1 white ball when 2 balls are randomly drawn, is 1/2. Which of the following is the ratio of the number of black balls to white balls in the bag

A. 1:7
B. 2:1
C. 1:1
D. 4:1
E. 1:4


Hi, I did it with another method, it took a lot of trial and error to find a way to solve this one. At first I took the pure algebraic way to solve it, but no joy. Then I kind of tweaked my method a bit. I have taken a method similar to chetan2u. As we have been asked the ratio of Black and White balls, let's call the ratio k.

Let the number of Black balls = kx
Let the number of White balls = x

Now let's write the probability expression

\(\frac{(Number of ways of selecting one white ball*Number of ways of selecting one black ball)}{Total number of ways of selecting any two balls}\)

\(\frac{(kx*x)}{(kx + x)(kx + x - 1)} = \frac{1}{2}\)

on simplifying we get

\(x^2(k^2 + 1 - 2k) = x(k + 1)\)

Here, we know that x(number of white balls) is positive and an integer. So we can simply cut x here

\(x(k^2 + 1 - 2k) = (k + 1)\)

\(x = \frac{(k + 1)}{(k^2 + 1 - 2k)}\)

Now, we know that x has to be an integer. From here, we can try values for k from options such that x turns out to be an integer.

Only (B) satisfies.
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Re: In a bag there are a certain number of black balls and white balls. Th  [#permalink]

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New post 05 Jul 2017, 23:35
ShashankDave wrote:
As we have been asked the ratio of Black and White balls, let's call the ratio k.

Let the number of Black balls = kx
Let the number of White balls = x

Now let's write the probability expression

\(\frac{(Number of ways of selecting one white ball*Number of ways of selecting one black ball)}{Total number of ways of selecting any two balls}\)

\(\frac{(kx*x)}{(kx + x)(kx + x - 1)} = \frac{1}{2}\)

on simplifying we get

\(x^2(k^2 + 1 - 2k) = x(k + 1)\)

Here, we know that x(number of white balls) is positive and an integer. So we can simply cut x here

\(x(k^2 + 1 - 2k) = (k + 1)\)

\(x = \frac{(k + 1)}{(k^2 + 1 - 2k)}\)

Now, we know that x has to be an integer. From here, we can try values for k from options such that x turns out to be an integer.

Only (B) satisfies.


Hi,

Your solution using variable is awesome! This algebraic way always excites me :P

However, I've just found out an error in a half-way step, thus making the final answer not as you expected.

It runs smoothly until this step:

\(\frac{(Number of ways of selecting one white ball*Number of ways of selecting one black ball)}{Total number of ways of selecting any two balls}\)

\(\frac{(kx*x)}{(kx + x)(kx + x - 1)} = \frac{1}{2}\)

In fact, after simplifying the equation, we will get the following:

\(x^2(k^2 + 1) = x(k + 1)\)
(In your post, you just made an extra \(-2kx^2\))

Moving on, because x(number of white balls) is a positive integer, we can simply cut x:

\(x(k^2 + 1) = (k + 1)\)

\(x = \frac{(k + 1)}{(k^2 + 1)}\)

x is a positive integer only when k=1. Therefore, the ratio of the number of black balls to white balls is 1:1

(C) must be the correct answer if the above probability equation (in red color) is correct!

Still not understand why OA is (B). Any thought?
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Re: In a bag there are a certain number of black balls and white balls. Th  [#permalink]

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New post 06 Jul 2017, 00:06
Lucy Phuong wrote:
ShashankDave wrote:
As we have been asked the ratio of Black and White balls, let's call the ratio k.

Let the number of Black balls = kx
Let the number of White balls = x

Now let's write the probability expression

\(\frac{(Number of ways of selecting one white ball*Number of ways of selecting one black ball)}{Total number of ways of selecting any two balls}\)

\(\frac{(kx*x)}{(kx + x)(kx + x - 1)} = \frac{1}{2}\)

on simplifying we get

\(x^2(k^2 + 1 - 2k) = x(k + 1)\)

Here, we know that x(number of white balls) is positive and an integer. So we can simply cut x here

\(x(k^2 + 1 - 2k) = (k + 1)\)

\(x = \frac{(k + 1)}{(k^2 + 1 - 2k)}\)

Now, we know that x has to be an integer. From here, we can try values for k from options such that x turns out to be an integer.

Only (B) satisfies.


Hi,

Your solution using variable is awesome! This algebraic way always excites me :P

However, I've just found out an error in a half-way step, thus making the final answer not as you expected.

It runs smoothly until this step:

\(\frac{(Number of ways of selecting one white ball*Number of ways of selecting one black ball)}{Total number of ways of selecting any two balls}\)

\(\frac{(kx*x)}{(kx + x)(kx + x - 1)} = \frac{1}{2}\)

In fact, after simplifying the equation, we will get the following:

\(x^2(k^2 + 1) = x(k + 1)\)
(In your post, you just made an extra \(-2kx^2\))

Moving on, because x(number of white balls) is a positive integer, we can simply cut x:

\(x(k^2 + 1) = (k + 1)\)

\(x = \frac{(k + 1)}{(k^2 + 1)}\)

x is a positive integer only when k=1. Therefore, the ratio of the number of black balls to white balls is 1:1

(C) must be the correct answer if the above probability equation (in red color) is correct!

Still not understand why OA is (B). Any thought?


In the highlighted part above LHS should be multiplied by 2 because we can have BW as well as WB cases. Having said that the question is not good. chetan is right, the ratio could be 6:3 (2:1) or 3:6 (1:2) among many other cases. So, you can ignore this question and move on.

TOPIC IS LOCKED.
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Re: In a bag there are a certain number of black balls and white balls. Th   [#permalink] 06 Jul 2017, 00:06
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