Question

A bag contains only red and blue marbles. Yasmine takes one marble at random from the bag. The probability that she takes a red marble is 1 in 5. Yasmine returns the marble to the bag and adds five more red marbles to the bag. The probability that she takes one red marble at random is now 1 in 3. How many red marbles were originally in the bag?

10 Points.

Answer 1

Answer: 5

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Step-by-step explanation:

Let
x = number of red marbles originally in the bag
y = number of blue marbles originally in the bag

The total number of marbles is x+y since there are only red marbles and blue marbles

The probability of picking red is 1/5, so we can form this equation
x/(x+y) = 1/5
which is the ratio of x to (x+y): the ratio of number of red to the number total

Cross multiply and isolate y
x/(x+y) = 1/5
5x = 1(x+y)
5x = x+y
y = 5x-x
y = 4x

Now that we add 5 more red marbles, we go from x red to (x+5) red. The total jumps from (x+y) to (x+y+5). The probability of picking red is now 1/3, so the equation is:
(x+5)/(x+y+5) = 1/3

Plug in y = 4x found earlier. Then solve for x
(x+5)/(x+y+5) = 1/3
(x+5)/(x+4x+5) = 1/3
(x+5)/(5x+5) = 1/3
3(x+5) = 1(5x+5)
3x+15 = 5x+5
5x-3x = 15-5
2x = 10
x = 5

If x = 5, then y is
y = 4x
y = 4*5
y = 20

Originally there are 5 red and 20 blue marbles making 5+20 = 25 total

Note how
P(red) = 5/25 = 1/5
for the first draw

If we added 5 more red then we jump to 10 red, 20 blue, 30 total
The probability of picking red is now
P(red) = 10/30 = 1/3
So both equations check out. The answer is confirmed.