# Thread: I solved this problem strangely?

1. ## I solved this problem strangely?

My problem is:
My first urn contains 10 balls, 4 red and 6 blue. My second one contains 16 red balls and some blue balls. A single ball is drawn from each urn. The probability that both balls are the same color is 0.44. How many blue balls are in the second urn.

I tried to muddle through it using P(A) and P(B) type notation, but I got nowhere. My solution is below:

I noticed that my odds of getting the same ball are on a straight line. With the X axis as the percentage of red balls in the second urn, and the Y axis as the percentage of balls being the same color, I got (0, 0.4), (50, 0.5), and (100, 0.6). So I just mentally plotted (20, 0.44). That gives me 20% of the urn 2 total being blue, for 4 balls.

I'm pretty sure this is correct since it's one of my choices, I'm just not sure this is the way it's supposed to be done. Is there a way to do this without even mental graphing? A way that uses P(A), P(A'), etc.?

2. Originally Posted by Boris B
My problem is:
My first urn contains 10 balls, 4 red and 6 blue. My second one contains 16 red balls and some blue balls. A single ball is drawn from each urn. The probability that both balls are the same color is 0.44. How many blue balls are in the second urn.

I tried to muddle through it using P(A) and P(B) type notation, but I got nowhere. My solution is below:

I noticed that my odds of getting the same ball are on a straight line. With the X axis as the percentage of red balls in the second urn, and the Y axis as the percentage of balls being the same color, I got (0, 0.4), (50, 0.5), and (100, 0.6). So I just mentally plotted (20, 0.44). That gives me 20% of the urn 2 total being blue, for 4 balls.

I'm pretty sure this is correct since it's one of my choices, I'm just not sure this is the way it's supposed to be done. Is there a way to do this without even mental graphing? A way that uses P(A), P(A'), etc.?
Let x be the number of blue balls in the second urn.

Then $\left (\frac{4}{10} \right) \left(\frac{16}{16+x}\right) + \left (\frac{6}{10}\right) \left(\frac{x}{16+x}\right) = 0.44$.

The solution to this equation is x = 4.

3. Hello, Boris!

My first urn contains 10 balls, 4 red and 6 blue.
My second one contains 16 red balls and some blue balls.
A single ball is drawn from each urn.
The probability that both balls are the same color is 0.44
How many blue balls are in the second urn?
Let $x$ = number of blue balls in Urn #2.

Urn 1 contains 10 balls of which 6 are blue.
. . $P(\text{Urn 1 }\wedge\text{ Blue}) \:=\:\frac{6}{10} \:=\:\frac{3}{5}$

Urn 2 contains $x + 16$ balls, of which $x$ are blue.
. . $P(\text{Urn 2 }\wedge\text{ Blue}) \:=\:\frac{x}{x+16}$

Hence: . $P(\text{both Blue}) \;=\;\frac{3}{5}\cdot\frac{x}{x+15} \;=\;0.44 \quad\Rightarrow\quad \frac{3x}{5(x+154)}\;=\;\frac{11}{25}$

Multiply by $25(x+16)\!:\quad15x \;=\;11(x+16)\quad\Rightarrow\quad x \:=\:44$

Therefore, there are 44 blue balls in Urn 2.

4. Originally Posted by Soroban
Hello, Boris!

Let $x$ = number of blue balls in Urn #2.

Urn 1 contains 10 balls of which 6 are blue.
. . $P(\text{Urn 1 }\wedge\text{ Blue}) \:=\:\frac{6}{10} \:=\:\frac{3}{5}$

Urn 2 contains $x + 16$ balls, of which $x$ are blue.
. . $P(\text{Urn 2 }\wedge\text{ Blue}) \:=\:\frac{x}{x+16}$

Hence: . $P(\text{both Blue}) \;=\;\frac{3}{5}\cdot\frac{x}{x+15} \;=\;0.44 \quad\Rightarrow\quad \frac{3x}{5(x+154)}\;=\;\frac{11}{25}$

Multiply by $25(x+16)\!:\quad15x \;=\;11(x+16)\quad\Rightarrow\quad x \:=\:44$

Therefore, there are 44 blue balls in Urn 2.

Soroban, both balls are the same colour so you can have BB or RR.

You've calculated from both balls being blue only .....

5. You're right, Mr. F . . . *slap head*

$P(\text{both blue}) \:=\:\frac{3}{5}\cdot\frac{x}{x+16}$

$P(\text{both red}) \:=\:\frac{2}{5}\cdot\frac{16}{x+16}$

$P(\text{same color}) \:=\:\frac{3x}{5(x+16)} + \frac{32}{5(x+16)} \:=\:\frac{11}{25} \quad\Rightarrow\quad\boxed{x \:=\:4}$

6. Hi, new to this forum!
i know my understanding of this problem is wrong and i've looked at Mr. Fantastic's and Soroban's explanations, but i must be missing an assumption thats obvious (but can't see it). can someone explain exactly why both probabilities need to be added together to eqaul .44? I'm reading it as since either p(b1) x p(b2) = 0.44 as does p(r1) x p(r2) =0.44, then all that needs to be done is take 6/10 * x/(x+16) = .44.

I know this is wrong (and that 4 is the correct answer) but i don't know why . Can someone break it down for me?

7. Originally Posted by uneakbreed
Hi, new to this forum!
i know my understanding of this problem is wrong and i've looked at Mr. Fantastic's and Soroban's explanations, but i must be missing an assumption thats obvious (but can't see it). can someone explain exactly why both probabilities need to be added together to eqaul .44? I'm reading it as since either p(b1) x p(b2) = 0.44 as does p(r1) x p(r2) =0.44, then all that needs to be done is take 6/10 * x/(x+16) = .44.

I know this is wrong (and that 4 is the correct answer) but i don't know why . Can someone break it down for me?
This is the probability of getting two blue balls: $\left (\frac{4}{10} \right) \left(\frac{16}{16+x}\right)$

This is the probability of getting two red balls: $\left (\frac{6}{10}\right) \left(\frac{x}{16+x}\right)$.

The probabilities added together is the probability of getting two balls of the same color, Blue/Blue or Red/Red.

Taking $\left (\frac{6}{10}\right) \left(\frac{x}{16+x}\right)$ would only give you the probability of getting two blue balls.
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