There are only green pens and red pens in a box. there are 3 more red pens than green pens in the box. sheila is going to take at random two pens from the box the probability that sheila will take two pens of the same color is 17/35 work out two different numbers of green pens that could be in the box

6 or 9

Step-by-step explanation:

Given the following :

Let the number of green pens = x

Number of red pens = x + 3

Probability of picking same color = 17/35

Taking two pens at random; probability of picking two pens of same color.

Probability of picking red on first pick then red on second pick ; or picking blue on first pick then blue on second pick

Probability = (Required outcome / Total possible outcomes)

Total number of pens = x + x + 3 = 2x + 3

Probability of picking red then red:

P(red first) = (x+3)/2x+3

P(red second) = x+3-1 / 2x+3-1 = (x+2)/2x+2)

Therefore, probability of red then red =

(x+3)/(2x+3) × (x+2)/2x+2)

= (x+3)(x+2) / (2x+3)(2x+2)

Probability of green then green:

P(first green) = x/(2x+3)

P(second green) = (x-1) / (2x+3-1) = (x-1) / (2x+2)

P(green then green) = x(x-1)/(2x+3)(2x+2)

Therefore,

[(x+3)(x+2) / (2x+3)(2x+2)] + [x(x-1)/(2x+3)(2x+2)] = 17/35

(x+3)(x+2)+x(x-1) / (2x+3)(2x+2) = 17/35

Cross multiply :

35(x+3)(x+2)+x(x-1) = 17(2x+3)(2x+2)

35(2x^2 + 4x + 6) = 17(4x^2 + 10x + 6)

70x^2 + 140x + 210 = 68x^2 + 170x + 102

70x^2 - 68x^2 + 140x - 170x + 210 - 102 = 0

2x^2 - 30x + 108 = 0

Now we have a quadratic equation which can be factoeized used using any known factorization method.

Factorizing this, we get

(x-6) = 0 or (x-9) = 0

x = 6 or x = 9