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Question
A bag contains 5 red, 4 blue and an unknown number m of green balls. If the probability of getting both the balls green, when two balls are selected at random is `1/7`, find m
Solution
The bag contains (4 + 5 + m) = (m + 9) balls.
If two balls are drawn from the bag, n(S) = (m+9)C2
Let G ≡ the event that both balls are green
There are m green balls
∴ 2 green balls can be drawn in mC2 ways.
∴ n(G) = mC2
It is given that,
P(G) = `1/7`
∴ `("n"("G"))/("n"("S")) = 1/7`
∴ 7mC2 = (m+9)C2
∴ `(7"m"("m" - 1))/(1*2) = (("m" + 9)("m" + 8))/(1*2)`
∴ 7m2 – 7m = m2 + 17m + 72
∴ 6m2 – 24m – 72 =0
∴ m2 – 4m – 12 =0
∴ (m – 6)(m + 2) = 0
∴ m = 6 or m= – 2
∴ m cannot be negative
∴ m = 6.
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