Mixing
Extracting two balls with the same color vs distinct color
$begingroup$
I am trying to find which has higher probability when extracting from a box without replacement: two balls of the same color, or two balls with different colors.
Let's assume that in the box are $a$ white balls and $b$ black balls.
$$P(text{distinct color})=P(text{first white})P(text{second black/first white})+P(text{first black})P(text{second white/first black})$$
$$=frac{a}{a+b}frac{b}{a+b-1}+frac{b}{a+b}frac{a}{a+b-1}=frac{2ab}{(a+b)(a+b-1)}$$
$$P(text{same color})=P(text{first white})P(text{second white/first white})+P(text{first black})P(text{second black/first black})$$
$$=frac{a}{a+b}frac{a-1}{a+b-1}+frac{b}{a+b}frac{b-1}{a+b-1}=frac{a^2+b^2-(a+b)}{(a+b)(a+b-1)}$$
But there is still needed to find which is more probably to happen, the case which I don't know how to proceed. Do I have to solve (I don't know how) for which domain of $a$ and $b$ does the following inequality hold true?
$$a^2+b^2-(a+b)< 2ab equiv (a-b)^2<a+b $$
This is like in the below answer, so I did it correctly, but still that doesn't help me to conclude an answer..
probability inequality
edited Feb 5 at 3:54
mjqxxxx
31.6k24086
$endgroup$
add a comment |
$begingroup$
I am trying to find which has higher probability when extracting from a box without replacement: two balls of the same color, or two balls with different colors.
Let's assume that in the box are $a$ white balls and $b$ black balls.
$$P(text{distinct color})=P(text{first white})P(text{second black/first white})+P(text{first black})P(text{second white/first black})$$
$$=frac{a}{a+b}frac{b}{a+b-1}+frac{b}{a+b}frac{a}{a+b-1}=frac{2ab}{(a+b)(a+b-1)}$$
$$P(text{same color})=P(text{first white})P(text{second white/first white})+P(text{first black})P(text{second black/first black})$$
$$=frac{a}{a+b}frac{a-1}{a+b-1}+frac{b}{a+b}frac{b-1}{a+b-1}=frac{a^2+b^2-(a+b)}{(a+b)(a+b-1)}$$
But there is still needed to find which is more probably to happen, the case which I don't know how to proceed. Do I have to solve (I don't know how) for which domain of $a$ and $b$ does the following inequality hold true?
$$a^2+b^2-(a+b)< 2ab equiv (a-b)^2<a+b $$
This is like in the below answer, so I did it correctly, but still that doesn't help me to conclude an answer..
probability inequality
edited Feb 5 at 3:54
mjqxxxx
31.6k24086
$endgroup$
1
$begingroup$
I have added a chart to my answer for visual reference.
$endgroup$
– Daniel Mathias
Feb 2 at 0:07
add a comment |
$begingroup$
I am trying to find which has higher probability when extracting from a box without replacement: two balls of the same color, or two balls with different colors.
Let's assume that in the box are $a$ white balls and $b$ black balls.
$$P(text{distinct color})=P(text{first white})P(text{second black/first white})+P(text{first black})P(text{second white/first black})$$
$$=frac{a}{a+b}frac{b}{a+b-1}+frac{b}{a+b}frac{a}{a+b-1}=frac{2ab}{(a+b)(a+b-1)}$$
$$P(text{same color})=P(text{first white})P(text{second white/first white})+P(text{first black})P(text{second black/first black})$$
$$=frac{a}{a+b}frac{a-1}{a+b-1}+frac{b}{a+b}frac{b-1}{a+b-1}=frac{a^2+b^2-(a+b)}{(a+b)(a+b-1)}$$
But there is still needed to find which is more probably to happen, the case which I don't know how to proceed. Do I have to solve (I don't know how) for which domain of $a$ and $b$ does the following inequality hold true?
$$a^2+b^2-(a+b)< 2ab equiv (a-b)^2<a+b $$
This is like in the below answer, so I did it correctly, but still that doesn't help me to conclude an answer..
probability inequality
edited Feb 5 at 3:54
mjqxxxx
31.6k24086
$endgroup$
I am trying to find which has higher probability when extracting from a box without replacement: two balls of the same color, or two balls with different colors.
Let's assume that in the box are $a$ white balls and $b$ black balls.
$$P(text{distinct color})=P(text{first white})P(text{second black/first white})+P(text{first black})P(text{second white/first black})$$
$$=frac{a}{a+b}frac{b}{a+b-1}+frac{b}{a+b}frac{a}{a+b-1}=frac{2ab}{(a+b)(a+b-1)}$$
$$P(text{same color})=P(text{first white})P(text{second white/first white})+P(text{first black})P(text{second black/first black})$$
$$=frac{a}{a+b}frac{a-1}{a+b-1}+frac{b}{a+b}frac{b-1}{a+b-1}=frac{a^2+b^2-(a+b)}{(a+b)(a+b-1)}$$
But there is still needed to find which is more probably to happen, the case which I don't know how to proceed. Do I have to solve (I don't know how) for which domain of $a$ and $b$ does the following inequality hold true?
$$a^2+b^2-(a+b)< 2ab equiv (a-b)^2<a+b $$
This is like in the below answer, so I did it correctly, but still that doesn't help me to conclude an answer..
probability inequality
probability inequality
edited Feb 5 at 3:54
mjqxxxx
31.6k24086
edited Feb 5 at 3:54
mjqxxxx
31.6k24086
edited Feb 5 at 3:54
mjqxxxx
31.6k24086
edited Feb 5 at 3:54
mjqxxxx
31.6k24086
edited Feb 5 at 3:54
mjqxxxx
31.6k24086
31.6k24086
asked Jan 26 at 0:32
user625055
1
$begingroup$
I have added a chart to my answer for visual reference.
$endgroup$
– Daniel Mathias
Feb 2 at 0:07
add a comment |
1
$begingroup$
I have added a chart to my answer for visual reference.
$endgroup$
– Daniel Mathias
Feb 2 at 0:07
1
1
$begingroup$
I have added a chart to my answer for visual reference.
$endgroup$
– Daniel Mathias
Feb 2 at 0:07
$begingroup$
I have added a chart to my answer for visual reference.
$endgroup$
– Daniel Mathias
Feb 2 at 0:07
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your inequality
begin{align*}
(a-b)^2< a + b
end{align*}
can be expressed as
begin{align*}
a - sqrt{2a + frac{1}{4}}+frac{1}{2} < b < a + sqrt{2a + frac{1}{4}}+frac{1}{2}
end{align*}
or
begin{align*}
left|b - a - frac{1}{2}right| < sqrt{2a + frac{1}{4}}
end{align*}
This domain isn't pretty in any way, but we can make statements about its behavior as $a,b rightarrow infty$. For example, dividing both sides by $a$,
begin{align*}
left|frac{b}{a} - 1 - frac{1}{2a}right| < sqrt{frac{2}{a} + frac{1}{4a^2}}
end{align*}
Therefore, in order for the probability of distinct colors to be greater than probability of the same, we need $b = a$ for sufficiently large values of $a$.
answered Jan 30 at 18:54
Tom ChenTom Chen
1,768715
$endgroup$
1
$begingroup$
correct (+1), and you may add that it is just the interior of the parabola $s=d^2$, where $s=a+b ;; d=a-b$
$endgroup$
– G Cab
Feb 2 at 1:49
add a comment |
$begingroup$
Total number of possible extractions is $binom{a+b}{2}$
Number of ways to extract:
- two white balls: $binom{a}{2}$
- two black balls: $binom{b}{2}$
- one of each color: $ab$
Probabilities can be calculated as $$frac{binom{a}{2}+binom{b}{2}}{binom{a+b}{2}}text{ for same color and }frac{ab}{binom{a+b}{2}}text{ for different colors}$$
Since you only want to know which has the higher probability, you can simply compare $binom{a}{2}+binom{b}{2}$ with $ab$ to see which is greater. Dependent on the values of $a$ and $b$, this could go either way.
Here's a chart detailing which outcome is more likely for $a,ble30$. A plus (+) indicates that one ball of each color is more likely, a minus (-) indicates that two balls of one color is more likely, and the equal (=) indicates that the probability is split 50-50. Notice that they are equal when $a$ and $b$ are consectutive triangular numbers (1,3,6,10,15...).
Generally, when there are approximately equal numbers of each color, then drawing one ball of each color is more likely.
1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0
1 + + = - - - - - - - - - - - - - - - - - - - - - - - - - - -
2 + + + + - - - - - - - - - - - - - - - - - - - - - - - - - -
3 = + + + + = - - - - - - - - - - - - - - - - - - - - - - - -
4 - + + + + + + - - - - - - - - - - - - - - - - - - - - - - -
5 - - + + + + + + - - - - - - - - - - - - - - - - - - - - - -
6 - - = + + + + + + = - - - - - - - - - - - - - - - - - - - -
7 - - - + + + + + + + + - - - - - - - - - - - - - - - - - - -
8 - - - - + + + + + + + + - - - - - - - - - - - - - - - - - -
9 - - - - - + + + + + + + + - - - - - - - - - - - - - - - - -
10 - - - - - = + + + + + + + + = - - - - - - - - - - - - - - -
11 - - - - - - + + + + + + + + + + - - - - - - - - - - - - - -
12 - - - - - - - + + + + + + + + + + - - - - - - - - - - - - -
13 - - - - - - - - + + + + + + + + + + - - - - - - - - - - - -
14 - - - - - - - - - + + + + + + + + + + - - - - - - - - - - -
15 - - - - - - - - - = + + + + + + + + + + = - - - - - - - - -
16 - - - - - - - - - - + + + + + + + + + + + + - - - - - - - -
17 - - - - - - - - - - - + + + + + + + + + + + + - - - - - - -
18 - - - - - - - - - - - - + + + + + + + + + + + + - - - - - -
19 - - - - - - - - - - - - - + + + + + + + + + + + + - - - - -
20 - - - - - - - - - - - - - - + + + + + + + + + + + + - - - -
21 - - - - - - - - - - - - - - = + + + + + + + + + + + + = - -
22 - - - - - - - - - - - - - - - + + + + + + + + + + + + + + -
23 - - - - - - - - - - - - - - - - + + + + + + + + + + + + + +
24 - - - - - - - - - - - - - - - - - + + + + + + + + + + + + +
25 - - - - - - - - - - - - - - - - - - + + + + + + + + + + + +
26 - - - - - - - - - - - - - - - - - - - + + + + + + + + + + +
27 - - - - - - - - - - - - - - - - - - - - + + + + + + + + + +
28 - - - - - - - - - - - - - - - - - - - - = + + + + + + + + +
29 - - - - - - - - - - - - - - - - - - - - - + + + + + + + + +
30 - - - - - - - - - - - - - - - - - - - - - - + + + + + + + +
answered Jan 26 at 2:52
Daniel MathiasDaniel Mathias
1,36018
$endgroup$
$begingroup$
You do realise the inequality you get is exactly what the OP got as well?
$endgroup$
– Macavity
Jan 26 at 8:16
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your inequality
begin{align*}
(a-b)^2< a + b
end{align*}
can be expressed as
begin{align*}
a - sqrt{2a + frac{1}{4}}+frac{1}{2} < b < a + sqrt{2a + frac{1}{4}}+frac{1}{2}
end{align*}
or
begin{align*}
left|b - a - frac{1}{2}right| < sqrt{2a + frac{1}{4}}
end{align*}
This domain isn't pretty in any way, but we can make statements about its behavior as $a,b rightarrow infty$. For example, dividing both sides by $a$,
begin{align*}
left|frac{b}{a} - 1 - frac{1}{2a}right| < sqrt{frac{2}{a} + frac{1}{4a^2}}
end{align*}
Therefore, in order for the probability of distinct colors to be greater than probability of the same, we need $b = a$ for sufficiently large values of $a$.
answered Jan 30 at 18:54
Tom ChenTom Chen
1,768715
$endgroup$
1
$begingroup$
correct (+1), and you may add that it is just the interior of the parabola $s=d^2$, where $s=a+b ;; d=a-b$
$endgroup$
– G Cab
Feb 2 at 1:49
add a comment |
$begingroup$
Your inequality
begin{align*}
(a-b)^2< a + b
end{align*}
can be expressed as
begin{align*}
a - sqrt{2a + frac{1}{4}}+frac{1}{2} < b < a + sqrt{2a + frac{1}{4}}+frac{1}{2}
end{align*}
or
begin{align*}
left|b - a - frac{1}{2}right| < sqrt{2a + frac{1}{4}}
end{align*}
This domain isn't pretty in any way, but we can make statements about its behavior as $a,b rightarrow infty$. For example, dividing both sides by $a$,
begin{align*}
left|frac{b}{a} - 1 - frac{1}{2a}right| < sqrt{frac{2}{a} + frac{1}{4a^2}}
end{align*}
Therefore, in order for the probability of distinct colors to be greater than probability of the same, we need $b = a$ for sufficiently large values of $a$.
answered Jan 30 at 18:54
Tom ChenTom Chen
1,768715
$endgroup$
1
$begingroup$
correct (+1), and you may add that it is just the interior of the parabola $s=d^2$, where $s=a+b ;; d=a-b$
$endgroup$
– G Cab
Feb 2 at 1:49
add a comment |
$begingroup$
Your inequality
begin{align*}
(a-b)^2< a + b
end{align*}
can be expressed as
begin{align*}
a - sqrt{2a + frac{1}{4}}+frac{1}{2} < b < a + sqrt{2a + frac{1}{4}}+frac{1}{2}
end{align*}
or
begin{align*}
left|b - a - frac{1}{2}right| < sqrt{2a + frac{1}{4}}
end{align*}
This domain isn't pretty in any way, but we can make statements about its behavior as $a,b rightarrow infty$. For example, dividing both sides by $a$,
begin{align*}
left|frac{b}{a} - 1 - frac{1}{2a}right| < sqrt{frac{2}{a} + frac{1}{4a^2}}
end{align*}
Therefore, in order for the probability of distinct colors to be greater than probability of the same, we need $b = a$ for sufficiently large values of $a$.
answered Jan 30 at 18:54
Tom ChenTom Chen
1,768715
$endgroup$
Your inequality
begin{align*}
(a-b)^2< a + b
end{align*}
can be expressed as
begin{align*}
a - sqrt{2a + frac{1}{4}}+frac{1}{2} < b < a + sqrt{2a + frac{1}{4}}+frac{1}{2}
end{align*}
or
begin{align*}
left|b - a - frac{1}{2}right| < sqrt{2a + frac{1}{4}}
end{align*}
This domain isn't pretty in any way, but we can make statements about its behavior as $a,b rightarrow infty$. For example, dividing both sides by $a$,
begin{align*}
left|frac{b}{a} - 1 - frac{1}{2a}right| < sqrt{frac{2}{a} + frac{1}{4a^2}}
end{align*}
Therefore, in order for the probability of distinct colors to be greater than probability of the same, we need $b = a$ for sufficiently large values of $a$.
answered Jan 30 at 18:54
Tom ChenTom Chen
1,768715
answered Jan 30 at 18:54
Tom ChenTom Chen
1,768715
answered Jan 30 at 18:54
Tom ChenTom Chen
1,768715
answered Jan 30 at 18:54
Tom ChenTom Chen
1,768715
1,768715
1
$begingroup$
correct (+1), and you may add that it is just the interior of the parabola $s=d^2$, where $s=a+b ;; d=a-b$
$endgroup$
– G Cab
Feb 2 at 1:49
add a comment |
1
$begingroup$
correct (+1), and you may add that it is just the interior of the parabola $s=d^2$, where $s=a+b ;; d=a-b$
$endgroup$
– G Cab
Feb 2 at 1:49
1
1
$begingroup$
correct (+1), and you may add that it is just the interior of the parabola $s=d^2$, where $s=a+b ;; d=a-b$
$endgroup$
– G Cab
Feb 2 at 1:49
$begingroup$
correct (+1), and you may add that it is just the interior of the parabola $s=d^2$, where $s=a+b ;; d=a-b$
$endgroup$
– G Cab
Feb 2 at 1:49
add a comment |
$begingroup$
Total number of possible extractions is $binom{a+b}{2}$
Number of ways to extract:
- two white balls: $binom{a}{2}$
- two black balls: $binom{b}{2}$
- one of each color: $ab$
Probabilities can be calculated as $$frac{binom{a}{2}+binom{b}{2}}{binom{a+b}{2}}text{ for same color and }frac{ab}{binom{a+b}{2}}text{ for different colors}$$
Since you only want to know which has the higher probability, you can simply compare $binom{a}{2}+binom{b}{2}$ with $ab$ to see which is greater. Dependent on the values of $a$ and $b$, this could go either way.
Here's a chart detailing which outcome is more likely for $a,ble30$. A plus (+) indicates that one ball of each color is more likely, a minus (-) indicates that two balls of one color is more likely, and the equal (=) indicates that the probability is split 50-50. Notice that they are equal when $a$ and $b$ are consectutive triangular numbers (1,3,6,10,15...).
Generally, when there are approximately equal numbers of each color, then drawing one ball of each color is more likely.
1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0
1 + + = - - - - - - - - - - - - - - - - - - - - - - - - - - -
2 + + + + - - - - - - - - - - - - - - - - - - - - - - - - - -
3 = + + + + = - - - - - - - - - - - - - - - - - - - - - - - -
4 - + + + + + + - - - - - - - - - - - - - - - - - - - - - - -
5 - - + + + + + + - - - - - - - - - - - - - - - - - - - - - -
6 - - = + + + + + + = - - - - - - - - - - - - - - - - - - - -
7 - - - + + + + + + + + - - - - - - - - - - - - - - - - - - -
8 - - - - + + + + + + + + - - - - - - - - - - - - - - - - - -
9 - - - - - + + + + + + + + - - - - - - - - - - - - - - - - -
10 - - - - - = + + + + + + + + = - - - - - - - - - - - - - - -
11 - - - - - - + + + + + + + + + + - - - - - - - - - - - - - -
12 - - - - - - - + + + + + + + + + + - - - - - - - - - - - - -
13 - - - - - - - - + + + + + + + + + + - - - - - - - - - - - -
14 - - - - - - - - - + + + + + + + + + + - - - - - - - - - - -
15 - - - - - - - - - = + + + + + + + + + + = - - - - - - - - -
16 - - - - - - - - - - + + + + + + + + + + + + - - - - - - - -
17 - - - - - - - - - - - + + + + + + + + + + + + - - - - - - -
18 - - - - - - - - - - - - + + + + + + + + + + + + - - - - - -
19 - - - - - - - - - - - - - + + + + + + + + + + + + - - - - -
20 - - - - - - - - - - - - - - + + + + + + + + + + + + - - - -
21 - - - - - - - - - - - - - - = + + + + + + + + + + + + = - -
22 - - - - - - - - - - - - - - - + + + + + + + + + + + + + + -
23 - - - - - - - - - - - - - - - - + + + + + + + + + + + + + +
24 - - - - - - - - - - - - - - - - - + + + + + + + + + + + + +
25 - - - - - - - - - - - - - - - - - - + + + + + + + + + + + +
26 - - - - - - - - - - - - - - - - - - - + + + + + + + + + + +
27 - - - - - - - - - - - - - - - - - - - - + + + + + + + + + +
28 - - - - - - - - - - - - - - - - - - - - = + + + + + + + + +
29 - - - - - - - - - - - - - - - - - - - - - + + + + + + + + +
30 - - - - - - - - - - - - - - - - - - - - - - + + + + + + + +
answered Jan 26 at 2:52
Daniel MathiasDaniel Mathias
1,36018
$endgroup$
$begingroup$
You do realise the inequality you get is exactly what the OP got as well?
$endgroup$
– Macavity
Jan 26 at 8:16
add a comment |
$begingroup$
Total number of possible extractions is $binom{a+b}{2}$
Number of ways to extract:
- two white balls: $binom{a}{2}$
- two black balls: $binom{b}{2}$
- one of each color: $ab$
Probabilities can be calculated as $$frac{binom{a}{2}+binom{b}{2}}{binom{a+b}{2}}text{ for same color and }frac{ab}{binom{a+b}{2}}text{ for different colors}$$
Since you only want to know which has the higher probability, you can simply compare $binom{a}{2}+binom{b}{2}$ with $ab$ to see which is greater. Dependent on the values of $a$ and $b$, this could go either way.
Here's a chart detailing which outcome is more likely for $a,ble30$. A plus (+) indicates that one ball of each color is more likely, a minus (-) indicates that two balls of one color is more likely, and the equal (=) indicates that the probability is split 50-50. Notice that they are equal when $a$ and $b$ are consectutive triangular numbers (1,3,6,10,15...).
Generally, when there are approximately equal numbers of each color, then drawing one ball of each color is more likely.
1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0
1 + + = - - - - - - - - - - - - - - - - - - - - - - - - - - -
2 + + + + - - - - - - - - - - - - - - - - - - - - - - - - - -
3 = + + + + = - - - - - - - - - - - - - - - - - - - - - - - -
4 - + + + + + + - - - - - - - - - - - - - - - - - - - - - - -
5 - - + + + + + + - - - - - - - - - - - - - - - - - - - - - -
6 - - = + + + + + + = - - - - - - - - - - - - - - - - - - - -
7 - - - + + + + + + + + - - - - - - - - - - - - - - - - - - -
8 - - - - + + + + + + + + - - - - - - - - - - - - - - - - - -
9 - - - - - + + + + + + + + - - - - - - - - - - - - - - - - -
10 - - - - - = + + + + + + + + = - - - - - - - - - - - - - - -
11 - - - - - - + + + + + + + + + + - - - - - - - - - - - - - -
12 - - - - - - - + + + + + + + + + + - - - - - - - - - - - - -
13 - - - - - - - - + + + + + + + + + + - - - - - - - - - - - -
14 - - - - - - - - - + + + + + + + + + + - - - - - - - - - - -
15 - - - - - - - - - = + + + + + + + + + + = - - - - - - - - -
16 - - - - - - - - - - + + + + + + + + + + + + - - - - - - - -
17 - - - - - - - - - - - + + + + + + + + + + + + - - - - - - -
18 - - - - - - - - - - - - + + + + + + + + + + + + - - - - - -
19 - - - - - - - - - - - - - + + + + + + + + + + + + - - - - -
20 - - - - - - - - - - - - - - + + + + + + + + + + + + - - - -
21 - - - - - - - - - - - - - - = + + + + + + + + + + + + = - -
22 - - - - - - - - - - - - - - - + + + + + + + + + + + + + + -
23 - - - - - - - - - - - - - - - - + + + + + + + + + + + + + +
24 - - - - - - - - - - - - - - - - - + + + + + + + + + + + + +
25 - - - - - - - - - - - - - - - - - - + + + + + + + + + + + +
26 - - - - - - - - - - - - - - - - - - - + + + + + + + + + + +
27 - - - - - - - - - - - - - - - - - - - - + + + + + + + + + +
28 - - - - - - - - - - - - - - - - - - - - = + + + + + + + + +
29 - - - - - - - - - - - - - - - - - - - - - + + + + + + + + +
30 - - - - - - - - - - - - - - - - - - - - - - + + + + + + + +
answered Jan 26 at 2:52
Daniel MathiasDaniel Mathias
1,36018
$endgroup$
$begingroup$
You do realise the inequality you get is exactly what the OP got as well?
$endgroup$
– Macavity
Jan 26 at 8:16
add a comment |
$begingroup$
Total number of possible extractions is $binom{a+b}{2}$
Number of ways to extract:
- two white balls: $binom{a}{2}$
- two black balls: $binom{b}{2}$
- one of each color: $ab$
Probabilities can be calculated as $$frac{binom{a}{2}+binom{b}{2}}{binom{a+b}{2}}text{ for same color and }frac{ab}{binom{a+b}{2}}text{ for different colors}$$
Since you only want to know which has the higher probability, you can simply compare $binom{a}{2}+binom{b}{2}$ with $ab$ to see which is greater. Dependent on the values of $a$ and $b$, this could go either way.
Here's a chart detailing which outcome is more likely for $a,ble30$. A plus (+) indicates that one ball of each color is more likely, a minus (-) indicates that two balls of one color is more likely, and the equal (=) indicates that the probability is split 50-50. Notice that they are equal when $a$ and $b$ are consectutive triangular numbers (1,3,6,10,15...).
Generally, when there are approximately equal numbers of each color, then drawing one ball of each color is more likely.
1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0
1 + + = - - - - - - - - - - - - - - - - - - - - - - - - - - -
2 + + + + - - - - - - - - - - - - - - - - - - - - - - - - - -
3 = + + + + = - - - - - - - - - - - - - - - - - - - - - - - -
4 - + + + + + + - - - - - - - - - - - - - - - - - - - - - - -
5 - - + + + + + + - - - - - - - - - - - - - - - - - - - - - -
6 - - = + + + + + + = - - - - - - - - - - - - - - - - - - - -
7 - - - + + + + + + + + - - - - - - - - - - - - - - - - - - -
8 - - - - + + + + + + + + - - - - - - - - - - - - - - - - - -
9 - - - - - + + + + + + + + - - - - - - - - - - - - - - - - -
10 - - - - - = + + + + + + + + = - - - - - - - - - - - - - - -
11 - - - - - - + + + + + + + + + + - - - - - - - - - - - - - -
12 - - - - - - - + + + + + + + + + + - - - - - - - - - - - - -
13 - - - - - - - - + + + + + + + + + + - - - - - - - - - - - -
14 - - - - - - - - - + + + + + + + + + + - - - - - - - - - - -
15 - - - - - - - - - = + + + + + + + + + + = - - - - - - - - -
16 - - - - - - - - - - + + + + + + + + + + + + - - - - - - - -
17 - - - - - - - - - - - + + + + + + + + + + + + - - - - - - -
18 - - - - - - - - - - - - + + + + + + + + + + + + - - - - - -
19 - - - - - - - - - - - - - + + + + + + + + + + + + - - - - -
20 - - - - - - - - - - - - - - + + + + + + + + + + + + - - - -
21 - - - - - - - - - - - - - - = + + + + + + + + + + + + = - -
22 - - - - - - - - - - - - - - - + + + + + + + + + + + + + + -
23 - - - - - - - - - - - - - - - - + + + + + + + + + + + + + +
24 - - - - - - - - - - - - - - - - - + + + + + + + + + + + + +
25 - - - - - - - - - - - - - - - - - - + + + + + + + + + + + +
26 - - - - - - - - - - - - - - - - - - - + + + + + + + + + + +
27 - - - - - - - - - - - - - - - - - - - - + + + + + + + + + +
28 - - - - - - - - - - - - - - - - - - - - = + + + + + + + + +
29 - - - - - - - - - - - - - - - - - - - - - + + + + + + + + +
30 - - - - - - - - - - - - - - - - - - - - - - + + + + + + + +
answered Jan 26 at 2:52
Daniel MathiasDaniel Mathias
1,36018
$endgroup$
Total number of possible extractions is $binom{a+b}{2}$
Number of ways to extract:
- two white balls: $binom{a}{2}$
- two black balls: $binom{b}{2}$
- one of each color: $ab$
Probabilities can be calculated as $$frac{binom{a}{2}+binom{b}{2}}{binom{a+b}{2}}text{ for same color and }frac{ab}{binom{a+b}{2}}text{ for different colors}$$
Since you only want to know which has the higher probability, you can simply compare $binom{a}{2}+binom{b}{2}$ with $ab$ to see which is greater. Dependent on the values of $a$ and $b$, this could go either way.
Here's a chart detailing which outcome is more likely for $a,ble30$. A plus (+) indicates that one ball of each color is more likely, a minus (-) indicates that two balls of one color is more likely, and the equal (=) indicates that the probability is split 50-50. Notice that they are equal when $a$ and $b$ are consectutive triangular numbers (1,3,6,10,15...).
Generally, when there are approximately equal numbers of each color, then drawing one ball of each color is more likely.
1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0 1 2 3 4 5 6 7 8 9 0
1 + + = - - - - - - - - - - - - - - - - - - - - - - - - - - -
2 + + + + - - - - - - - - - - - - - - - - - - - - - - - - - -
3 = + + + + = - - - - - - - - - - - - - - - - - - - - - - - -
4 - + + + + + + - - - - - - - - - - - - - - - - - - - - - - -
5 - - + + + + + + - - - - - - - - - - - - - - - - - - - - - -
6 - - = + + + + + + = - - - - - - - - - - - - - - - - - - - -
7 - - - + + + + + + + + - - - - - - - - - - - - - - - - - - -
8 - - - - + + + + + + + + - - - - - - - - - - - - - - - - - -
9 - - - - - + + + + + + + + - - - - - - - - - - - - - - - - -
10 - - - - - = + + + + + + + + = - - - - - - - - - - - - - - -
11 - - - - - - + + + + + + + + + + - - - - - - - - - - - - - -
12 - - - - - - - + + + + + + + + + + - - - - - - - - - - - - -
13 - - - - - - - - + + + + + + + + + + - - - - - - - - - - - -
14 - - - - - - - - - + + + + + + + + + + - - - - - - - - - - -
15 - - - - - - - - - = + + + + + + + + + + = - - - - - - - - -
16 - - - - - - - - - - + + + + + + + + + + + + - - - - - - - -
17 - - - - - - - - - - - + + + + + + + + + + + + - - - - - - -
18 - - - - - - - - - - - - + + + + + + + + + + + + - - - - - -
19 - - - - - - - - - - - - - + + + + + + + + + + + + - - - - -
20 - - - - - - - - - - - - - - + + + + + + + + + + + + - - - -
21 - - - - - - - - - - - - - - = + + + + + + + + + + + + = - -
22 - - - - - - - - - - - - - - - + + + + + + + + + + + + + + -
23 - - - - - - - - - - - - - - - - + + + + + + + + + + + + + +
24 - - - - - - - - - - - - - - - - - + + + + + + + + + + + + +
25 - - - - - - - - - - - - - - - - - - + + + + + + + + + + + +
26 - - - - - - - - - - - - - - - - - - - + + + + + + + + + + +
27 - - - - - - - - - - - - - - - - - - - - + + + + + + + + + +
28 - - - - - - - - - - - - - - - - - - - - = + + + + + + + + +
29 - - - - - - - - - - - - - - - - - - - - - + + + + + + + + +
30 - - - - - - - - - - - - - - - - - - - - - - + + + + + + + +
answered Jan 26 at 2:52
Daniel MathiasDaniel Mathias
1,36018
edited Feb 2 at 0:09
answered Jan 26 at 2:52
Daniel MathiasDaniel Mathias
1,36018
answered Jan 26 at 2:52
Daniel MathiasDaniel Mathias
1,36018
answered Jan 26 at 2:52
Daniel MathiasDaniel Mathias
1,36018
1,36018
$begingroup$
You do realise the inequality you get is exactly what the OP got as well?
$endgroup$
– Macavity
Jan 26 at 8:16
add a comment |
$begingroup$
You do realise the inequality you get is exactly what the OP got as well?
$endgroup$
– Macavity
Jan 26 at 8:16
$begingroup$
You do realise the inequality you get is exactly what the OP got as well?
$endgroup$
– Macavity
Jan 26 at 8:16
$begingroup$
You do realise the inequality you get is exactly what the OP got as well?
$endgroup$
– Macavity
Jan 26 at 8:16
add a comment |
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$begingroup$
I have added a chart to my answer for visual reference.
$endgroup$
– Daniel Mathias
Feb 2 at 0:07