# Minimum number of red balls

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## Homework Statement

A bag contains n identical red balls, 2n identical black balls and 3n identical while balls. If probability of drawing n balls of same color is greater than or equal to 1/6, then minimum number of red balls in the bag is equal to?

## The Attempt at a Solution

$P = \dfrac{^nC_n+^{2n}C_n+^{3n}C_n}{^{6n}C_n} \geq \dfrac{1}{6}$

But it seems too difficult to solve this equation.

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Ray Vickson
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## Homework Statement

A bag contains n identical red balls, 2n identical black balls and 3n identical while balls. If probability of drawing n balls of same color is greater than or equal to 1/6, then minimum number of red balls in the bag is equal to?

## The Attempt at a Solution

$P = \dfrac{^nC_n+^{2n}C_n+^{3n}C_n}{^{6n}C_n} \geq \dfrac{1}{6}$

But it seems too difficult to solve this equation.
So, what have you tried so far?

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So, what have you tried so far?
Plugging in integral values starting from n=1(I know this sounds stupid, but I'm helpless). The inequality is satisfied by n=1 which seems obvious, though it is incorrect.

SammyS
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Plugging in integral values starting from n=1(I know this sounds stupid, but I'm helpless). The inequality is satisfied by n=1 which seems obvious, though it is incorrect.
Of course if n = 1, you will always draw one ball of the same color in a row !

Ray Vickson
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Plugging in integral values starting from n=1(I know this sounds stupid, but I'm helpless). The inequality is satisfied by n=1 which seems obvious, though it is incorrect.
It is not really incorrect, although you could argue that it does not meet the 'spirit' of the problem. What about some higher values of n? Do they work?

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It is not really incorrect, although you could argue that it does not meet the 'spirit' of the problem. What about some higher values of n? Do they work?
The correct answer is n=5 but it does not satisfy my equation.

Ray Vickson
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The correct answer is n=5 but it does not satisfy my equation.
How can you say n=5 is correct, but at the same time does not satisfy your equation? The only way that can happen is for your equation to be wrong. Who told you that n = 5 is correct?

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How can you say n=5 is correct, but at the same time does not satisfy your equation? The only way that can happen is for your equation to be wrong. Who told you that n = 5 is correct?
This question is from a test paper and the solution mentions that the answer is 5.
It also mentions that the required probability is $\dfrac{3}{^{n+2}C_2}$. This is where I'm confused. How did they arrive at this peculiar expression?

Ray Vickson
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This question is from a test paper and the solution mentions that the answer is 5.
It also mentions that the required probability is $\dfrac{3}{^{n+2}C_2}$. This is where I'm confused. How did they arrive at this peculiar expression?
I have no idea how they arrived at that incorrect expression; your answer is correct.

In fact, the answer sheet is even incorrect about the numerics: if you substitute n = 5 into their expression, you get a result < 1/6. The value n = 4 is the largest integer which makes their expression ≥ 1/6. In fact, n = 5 is the smallest integer that makes their expression ≤ 1/6. Maybe they meant "≤ 1/6" in the problem statement?

The value n = 2 is the largest integer that makes your expression ≥ 1/6, and n = 1 is the smallest.

haruspex
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Wrt how to solve the equation in the OP in general, it's clear that the 3nCn is the dominant term in the numerator. So throw away the other two terms and solve that using Stirling's formula. You should get an expression like (3355/2266)n >= 1/6. Then plug in successively lower n in the full equation until you find the boundary.
Of course, since it turns out that the question is flawed and the answer comes out as around n=1 or n=2, that approach is not going to help.

Ray Vickson
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Wrt how to solve the equation in the OP in general, it's clear that the 3nCn is the dominant term in the numerator. So throw away the other two terms and solve that using Stirling's formula. You should get an expression like (3355/2266)n >= 1/6. Then plug in successively lower n in the full equation until you find the boundary.
Of course, since it turns out that the question is flawed and the answer comes out as around n=1 or n=2, that approach is not going to help.

He could also go to the Wolfram Alpha site and enter
'plot (1 + binomial(2*n,n) + binomial(3*n,n))/binomial(6*n,n) for n from 1 to 10'
or---to get individual values---enter
'(1 + binomial(2*n,n) + binomial(3*n,n))/binomial(6*n,n) for n = 3' (or for n = 4 or 5 or 6 ...)

In probability problems, sometimes it is easier to answer the negation of the question then subtract the result from 1.

Ray Vickson