Question

You have a bag which contains only red and green marbles. In this bag with $x^{2} 5$ marbles total, $x 1$ are red. Also, $x-3$ marbles have a scratch on them. The probability of drawing a red marble from the original bag is equal to that of drawing a marble with a scratch from the marbles left in the bag after twenty scratch-free marbles are taken out of the full bag. How many marbles were originally in the bag?

Answer 1

The original number of marbles in the bag is 30.

The given parameters are:

$Red = x + 1$

$Total = x^2 + 5$

The probability of red is:

$P(Red) = \frac{Red}{Total}$

$P(Red) = \frac{x+1}{x^2 + 5}$

When 20 marbles were removed, the marbles left are:

$Marbles =Total - 20$

This gives:

$Marbles =x^2 + 5 - 20$

$Marbles =x^2 -15$

The probability of selecting a scratched marble at this point is:

$P(Scratch) = \frac{Scratch}{Marbles}$

$P(Scratch) = \frac{x -3}{x^2 - 15}$

This probability equals the probability of red.

i.e.

$P(Red) = P(Scratch)$

So, we have:

$\frac{x +1}{x^2 + 5} = \frac{x -3}{x^2 - 15}$

Cross multiply

$(x + 1)(x^2 - 15) = (x - 3)(x^2 + 5)$

Expand

$x^3 - 15x + x^2 - 15 = x^3 + 5x - 3x^2 - 15$

Subtract $x^3$ from both sides

$- 15x + x^2 - 15 = 5x - 3x^2 - 15$

Add 15 to both sides

$- 15x + x^2 = 5x - 3x^2$

Collect like terms

$3x^2 + x^2 - 15x - 5x = 0$

$4x^2 - 20x = 0$

Divide through by 4

$x^2 - 5x = 0$

Expand

$x(x - 5) = 0$

Split

$x = 0$ or $x - 5 = 0$

$x = 0$ or $x = 5$

x can't be 0.

So: $x = 5$

The number of marbles initially is:

$Total = x^2 + 5$

$Total = 5^2 + 5$

$Total = 30$

Hence, the original number of marbles in the bag is 30.

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