(1/3) * (5/113) * (x/y) = 0.05

2/3 + ((1/3) * (5/113) * ((y-x)/y)) = 0.63

Can anyone please help me out? Thanks! (Worried)

- Thread starter Dufus
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(1/3) * (5/113) * (x/y) = 0.05

2/3 + ((1/3) * (5/113) * ((y-x)/y)) = 0.63

Can anyone please help me out? Thanks! (Worried)

Is this a set of equations you need to solve simultaneously?

(1/3) * (5/113) * (x/y) = 0.05

2/3 + ((1/3) * (5/113) * ((y-x)/y)) = 0.63

Can anyone please help me out? Thanks! (Worried)

Yes, x is the same in both equations and y is the same in both equations. (Nod)Is this a set of equations you need to solve simultaneously?

HINT: above results in 100x = 339y(1/3) * (5/113) * (x/y) = 0.05

I appreciate the helpful nudge, but I'm getting more and more bogged down. Here's the rabbit hole I've followed from that:HINT: above results in 100x = 339y

100x = 339y

x = 3.39y

(1/3) * (5/113) * (3.39) = 0.05

2/3 + ((1/3) * (5/113) * ((y-(3.39*y))/y)) = 0.63

((1/3) * (5/113) * (3.39)) - (2/3) = -0.61666666666666666666666666666667

(1/3) * (5/113) * ((y-(3.39*y))/y) = -0.036666666666666666666666666666667

(((1/3) * (5/113) * (3.39)) - (2/3)) / (1/3) = -1.850000000000000000000000000001

(5/113) * ((y-(3.39*y))/y) = -0.11

(((1/3) * (5/113) * (3.39)) - (2/3)) / (1/3) / (5/113) = -41.810000000000000000000000000023

(y-(3.39*y))/y = -2.486

I've come unstuck at this point. (Rain)

I

x = 3.39y

(y-x)/y = (0.63-(2/3)) / (1/3) / (5/113)

I just don't know how to untangle that "(y-x)/y" beyond substituting x:

(y-(3.39*y))/y = (0.63-(2/3)) / (1/3) / (5/113)

And that doesn't seem to help much, assuming I'm even on the right track.

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Who created this problem? . . . What a horrible way to write it!

And don't use decimals!

Besides, the problem has

Solve the system: . \(\displaystyle \begin{array}{ccccc}\dfrac{1}{3}\!\cdot\!\dfrac{5}{113}\!\cdot\!\dfrac{x}{y} &=& \dfrac{1}{20} & {\color{blue}[1]} \\ \\[-3mm]

\dfrac{2}{3} + \dfrac{1}{3}\!\cdot\!\dfrac{5}{113}\!\cdot\!\dfrac{y-x}{y} &=& \dfrac{63}{100} & {\color{blue}[2]} \end{array}\)

As Wilmer pointed out, [1] becomes: .\(\displaystyle 100x \:=\:339y \quad\Rightarrow\quad \frac{x}{y} \:=\:\frac{339}{100}\) .(a)

[2] simplifies to: .\(\displaystyle \frac{2}{3} + \frac{5}{339}\cdot\frac{y-x}{y} \:=\:\frac{63}{100} \)

. . . . . . . . . . . \(\displaystyle \frac{2}{3} + \frac{5}{339}\left(1 - {\color{red}\frac{x}{y}}\right) \;=\;\frac{63}{100}\)

Substitute (a): . \(\displaystyle \frac{2}{3} + \frac{5}{339}\left(1 - {\color{red}\frac{339}{100}}\right) \;=\;\frac{63}{100}\)

. . . . . . . . . . . . . \(\displaystyle \frac{2}{3} + \frac{5}{339}\left(-\frac{239}{100}\right) \;=\;\frac{63}{100}\)

. . . . . . . . . . . . . . . . .\(\displaystyle \frac{2}{3} - \frac{1196}{33,900} \;=\;\frac{63}{100}\)

. . . . . . . . . . . . . . . . . . . \(\displaystyle \frac{21,405}{33,900} \;=\;\frac{63}{100}\)

. . And we have: . . . . . . . . \(\displaystyle \frac{1427}{2260} \;{\color{red}\neq}\;\frac{63}{100}\)

The system is inconsistent . . . It has

That'll be my fault. It actually began life as a computer programming problem, but I thought I'd found the best way to represent it in mathematical terms. Evidently not, sorry!Who created this problem? . . . What a horrible way to write it!

The system is inconsistent . . . It hassolution.no

Rather than go into the programming code, I'll explain the problem like this:

There are 113 marbles in a jar. Five of them are red, the rest are blue.

There's a one in three chance you'll be selected to play the game.

If you are, then blindfolded, you shake the jar and draw one marble completely at random.

If it's blue, you keep it and the game is over.

If it's red, there's an [x/y] chance I'll let you keep it, or a [1-(x/y)] chance I'll make you put it back. Whether you get to keep it or not, the game is over.

I want there to be a 5% probability of you drawing a red marble and keeping it. So what should x and y be?

Probem is the 2nd equation should be:(1/3) * (5/113) * (x/y) = 0.05

2/3 + ((1/3) * (5/113) * ((y-x)/y)) = 0.63

2/3 + (1/3) * (5/113) * ((y-x)/y) = 1427/2260

1427/2260 = .63141..... ; that was "rounded" to .63 : a NO-NO!

And since we have (from 1st equation) x = 3.39y,

then ANY value can be assigned to y!

I must be missing something: but that makes no sense to me.There are 113 marbles in a jar. Five of them are red, the rest are blue.

There's a one in three chance you'll be selected to play the game.

If you are, then blindfolded, you shake the jar and draw one marble completely at random.

If it's blue, you keep it and the game is over.

If it's red, there's an [x/y] chance I'll let you keep it, or a [1-(x/y)] chance I'll make you put it back. Whether you get to keep it or not, the game is over.

I want there to be a 5% probability of you drawing a red marble and keeping it. So what should x and y be?

The probability of drawing a red is 5/113 ; that's a ~4.42% probability.

Whatever you do AFTER the draw will certainly not INCREASE that.

If you said "I want there to be a 3% probability.....", then we have:

(let k = x/y)

5k/113 = 3/100

k = 339/500

Wilmer is absolutely correct.

There is still something terribly wrong with the problem . . .

To get a red marble and keep it, three events must occur:There are 113 marbles in a jar; 5 are red, the rest blue.

There's a one-in-three chance you'll be selected to play the game.

If you are, you draw one marble completely at random.

If it's blue, you keep it and the game is over.

If it's red, there's an \(\displaystyle \tfrac{x}{y}\) chance that I'll let you keep it,

. . and a \(\displaystyle 1 -\tfrac{x}{y}\) chance I'll make you put it back.

Whether you get to keep it or not, the game is over.

I want there to be a 5% probability of you drawing a red marble and keeping it.

So what should \(\displaystyle x\) and \(\displaystyle y\) be?

. . \(\displaystyle \text{[1] You get to play: }\;P(\text{play}) \:=\:\frac{1}{3}\)

. . \(\displaystyle \text{[2] You draw a red marble: }\;P(\text{red}) \:=\:\frac{5}{113}\)

. . \(\displaystyle \text{[3] You get to keep it: }\;P(\text{keep}) \:=\:\frac{x}{y}\)

\(\displaystyle \text{Hence: }\;P(\text{play} \wedge \text{red} \wedge \text{keep}) \;=\;\frac{1}{3}\cdot\frac{5}{113}\cdot\frac{x}{y}\)

. . This is to be .\(\displaystyle 5\% \,=\,\frac{1}{20}\)

Hence, we have: .\(\displaystyle \frac{5x}{339y} \:=\:\frac{1}{20} \quad\Rightarrow\quad \frac{x}{y} \:=\:\frac{339}{100} \;=\;339\% \) .??

. . A probability cannot be greater than 100%.