A bag contains red balls and white balls. If five balls are to be pulled from the bag, with replacement, the probability of getting exactly three red balls is 32 times the probability of getting exactly one red ball. What percent of the balls originally in the bag are red?

Percentage of balls which are red = 80%.

Step-by-step explanation:

Let the probability of drawing a red ball be x then the probability of drawing a white ball is 1-x.

There are 5C3 = 10 ways of getting 3 reds in 5 draws so the probability of this is 10 * x^3 * (1 - x) ^2.

The probability of getting 1 red in 5 draws = 5C1 * x (1-x) ^4.

The first probability is 32 times the last.

So we have the equation:

10x^3 (1 - x) ^2 / 5x (1 - x) ^4 = 32

2x^2 / (1 - x) ^2 = 32

2x^2 = 32 (1 - x) ^2

2x^2 = 32 (1 - 2x + x^2)

2x^2 = 32 - 64x + 32x^2

30x^2 - 64x + 32 = 0

15x^2 - 32x + 16 = 0

(5x - 4) (3x - 4) = 0

x = 0.8, 1.333 ...

The probability must be < 1 so x = 0.8 = 80%.