Question

A bag contains 5 red , 4 blue and an unknown number of x green balls. Two balls are
drawn at random. If the probability of both being green is $\frac{1}{7}$ , find x. Also find the probability of getting at least one green ball.

Answer 1

Answer:

Total number of balls = No.of red balls + No.of blue balls + No.of green balls

= 5+4+x = x+9

Now,

Probability of getting both as green balls

= (x/x+9)(x-1/x+8)

So,

x(x-1) / (x+9)(x+8) = 1/7

Or,

x^2 + 17x + 72 = 7(x^2 - x)

Or,

6x^2 - 24x - 72 = 0

x^2 - 4x - 12 = 0

(x-6)(x+2) = 0

Therefore, no

of green balls = 6 and not -2 since it cannot be -2.

Therefore,

Probability of getting at least one green ball = 6/15*(1-5/14) + 6/15*5/14 = 6/15 = 2/5

Hope this helps you !