Chips Probability

[linaxlovex13]

A bag has some number of chips of two different colors. Imagine picking out two chips from the bag. How many chips of each color should be in the bag so that there is a 50% chance of picking two chips of the same color, and a 50% chance of picking 2 chips of different colors?

>>>At first, I thought it might be four chips, but the probability didn't work out right so i'm pretty stuck
>>>Thank you in advance for your help!

 

[mmm4444bot]

Hmmm. At first glance, I'm wondering if this exercise has a valid answer.

But first, can you confirm that the events occur "without replacement"?

Also, what have you learned so far about determining probabilities?

You could also start naming things, so that we can discuss this senario.

For example, call the chips black and white.

b = the number of black chips

w = the number of white chips

Do you understand why the probability of first picking a black chip is b/(b + w)?

Now, is the black chip returned to the bag before the second pick?

If not, do you understand why (b - 1)/(b - 1 + w) is the probability of picking a second black chip?

Have you seen examples of when to multiply probabilites? When to add them?

Once we have some sense of where you're at, both within this exercise and within the subject in general, we'll be able to determine an appropriate place to begin helping you.

Cheers

 

[soroban]

Hello, linaxlovex13!

I have a solution, but it took Olympic-level gymnastics . . .

A bag has some number of chips of two different colors.
Imagine picking out two chips from the bag.
How many chips of each color should be in the bag so that there is a 50% chance of picking two chips of the same color
and a 50% chance of picking 2 chips of different colors?

\(\displaystyle \text{Let }B\text{ = number of black chips, }\:W\text{ = number of white chips.}\)

\(\displaystyle \text{Same color: }\:{B\choose2} + {W\choose2} \;=\;\frac{B(B-1)}{2} + \frac{W(W-1)}{2}\, \text{ ways.}\)

\(\displaystyle \text{Opposite colors: }\:{B\choose1}\cdot{W\choose1} \:=\:B\cdot W\,\text{ ways.}\)


\(\displaystyle \text{The numbers of ways are equal: }\;\frac{B(B-1)}{2} + \frac{W(W-1)}{2} \;=\;B\cdot W\)

. . \(\displaystyle \text{which simplifies to: }\;B^2 - (2W+1)B + W(W-1) \;=\;0\)

\(\displaystyle \text{Quadratic Formula: }\;B \;=\;\frac{(2W+1) \pm\sqrt{(2W+1)^2 - 4W(W-1)}}{2} \;=\;\frac{(2W+1) \pm\sqrt{8W+1}}{2}\)


\(\displaystyle \text{Since }B\text{ is an integer, }\:8W+1\text{ must be a square.}\)

\(\displaystyle \text{It is fairly well-known that }(8W+1)\text{ is a square if }W\text{ is a triangular number.}\)
. . \(\displaystyle \text{That is: }\:\boxed{W \:=\:\frac{n(n+1)}{2}}\)

\(\displaystyle \text{Substitute: }\;B \;=\;\frac{\left[2\,\frac{n(n+1)}{2} + 1\right] \pm\sqrt{8\,\frac{n(n+1)}{2} + 1}}{2} \;=\;\frac{(n^2+n+1) \pm(2n+1)}{2}\)

. . \(\displaystyle \text{which simplifies to: }\;B \;=\;\begin{Bmatrix}\dfrac{(n+1)(n+2)}{2} \\ \\[-4mm] \dfrac{n(n-1)}{2} \end{Bmatrix}\)


\(\displaystyle \text{It seems that we have }two\text{ solutions: }\;\begin{array}{cccc} \hline \\[-3mm] B \:=\:\frac{(n+1)(n+2)}{2} & & W \:=\:\frac{n(n+1)}{2} \\ \\[-3mm] \hline \\[-3mm] B \:=\:\frac{n(n-1)}{2} & & W \:=\:\frac{n(n+1)}{2}\\ \\[-3mm] \hline \end{array}\)


\(\displaystyle \text{Upon examination, we see that }B\text{ and }W\text{ are }two\;consecutive\;triangular\;numbers\text{ (in some order).}\)



\(\displaystyle \text{The triangular numbers are: }\;1, 3, 6, 10, 15, 21, 28, \hdots\)

\(\displaystyle \text{Disregarding the 1, use any two consecutive numbers for }B\text{ and }W.\)


~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

\(\displaystyle \text{For example: }\:B \:=\:10,\:W = 15\)

\(\displaystyle \text{There are: }{25\choose2} \:=\:300\text{ possible outcomes.}\)

. . \(\displaystyle \text{Same color: }\:{10\choose2} + {15\choose2} \:=\:45 + 105 \:=\:150\text{ ways.}\)

. . \(\displaystyle \text{Diff. colors: }\:{10\choose1}{15\choose1} \:=\:10\cdot15 \:=\: 150\text{ ways}\)

\(\displaystyle \text{Therefore: }\;\begin{array}{ccccc}P(\text{same}) &=& \frac{150}{300} &=& 50\% \\ \\[-3mm] P(\text{diff.}) &=& \frac{150}{300} &=& 50\% \end{array}\)

 

[galactus]

Very nice, Soroban. I like. Clever, yet simplistic.

 

[Deleted member 4993]

A bag has some number of chips of two different colors. Imagine picking out two chips from the bag. How many chips of each color should be in the bag so that there is a 50% chance of picking two chips of the same color, and a 50% chance of picking 2 chips of different colors?

>>>At first, I thought it might be four chips,

That is a correct answer - only that the number of chips in each color are not equal (1 & 3)

but the probability didn't work out right so i'm pretty stuck
>>>Thank you in advance for your help!