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A bag contains 5 Red, 4 Blue and x green balls. Two balls are drawn at random from the bag. If the probability of both being green is 1/7, find x. I am clear with the proceedings but stuck at solving equation of factorials, so please show detailed steps.

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    Can you add the equations you created to get the question? – 2017-02-08

2 Answers 2

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Total balls = x + 9

Green balls = x

Probability of 2 balls both green $$= \frac{\binom x2}{\binom{x + 9}{2}}$$

$$\frac 17= \frac{\frac{x!}{2! × (x - 2)!}}{\frac{(x+9)!}{2! × (x + 7)!}}$$

$\frac 17= \frac{x!}{2! × (x - 2)!} × \frac{2! × (x + 7)!}{(x+9)!}$

$\frac 17= \frac{x!}{(x - 2)!} × \frac{(x+7)!}{(x+9)!}$

$\frac 17= \frac{x \cdot (x-1) \cdot (x-2)!}{(x - 2)!} × \frac{(x + 7)!}{(x+9)(x+8)(x+7)!}$

$\frac 17= \frac {x \cdot (x-1)}{(x+9) \cdot (x+8)}$

$x^2 + 17x + 72 = 7x^2 - 7x$

$6x^2 - 24x -72 = 0$

$x^2 - 4x -12 = 0$

$x^2 -6x +2x - 12 = 0$

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

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

Either x = -2 or x = 6.

But x can't be negative.

So Green balls x = 6

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    In the 3rd step the last part should have been reciprocated because of the change in sign from ÷ to multiplication. – 2017-02-08
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    And please show further workings. – 2017-02-08
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    Sorry my mistake. I am solving it on my mobile. So little problem. – 2017-02-08
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    Done. If any doubt please do ask. – 2017-02-08
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    No doubts at all! Thank u – 2017-02-08
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    Mine pleasure :-) – 2017-02-08
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Think about the probablities of each independent draw: $ \frac{x}{5+4+x} \frac{x-1}{5+4+(x-1)}=\frac{1}{7}$

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    Okay. It was quite easy with this method but can you please solve using combination method? – 2017-02-08
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    Though I got what I wanted in other answer but I learnt a new method from your answer. Thank you. – 2017-02-08