ATQ, aC2/(a+ 4)C2 = 1/3 Or, {a(a – 1)}/{(a + 4)(a + 3)} = 1/3 Or, 3a2 – 3a = a2 + 4a + 3a + 12 Or, 2a2 – 10a – 12 = 0 Or, 2a2 – 12a + 2a – 12 = 0 Or, 2a(a – 6) + 2(a – 6) = 0 Or, (a – 6)(2a + 2) = 0 Since, the number of coins cannot be negative, therefore a = 6 Number of one rupees coins = a + 4 = 10 Number of two rupees coins = 2a = 12 Total number of coins = 10 + 12 = 22 Required probability = 12/22 = 6/11
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