Sum

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

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#### Solution

The bag contains (4 + 5 + m) = (m + 9) balls.

If two balls are drawn from the bag, n(S) = ^{(m+9)}C_{2}

Let G ≡ the event that both balls are green

There are m green balls

∴ 2 green balls can be drawn in ^{m}C_{2} ways.

∴ n(G) = ^{m}C_{2}

It is given that,

P(G) = `1/7`

∴ `("n"("G"))/("n"("S")) = 1/7`

∴ 7^{m}C_{2} = ^{(m+9)}C_{2}

∴ `(7"m"("m" - 1))/(1*2) = (("m" + 9)("m" + 8))/(1*2)`

∴ 7m^{2} – 7m = m^{2} + 17m + 72

∴ 6m^{2} – 24m – 72 =0

∴ m^{2} – 4m – 12 =0

∴ (m – 6)(m + 2) = 0

∴ m = 6 or m= – 2

∴ m cannot be negative

∴ m = 6.

Concept: Concept of Probability

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