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?answer it and get 98 points if you get it right
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.