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    Question

    A bag contains a few red balls and 2 green balls. Two

    balls are drawn randomly without replacement. The probability that both balls are red is (2/5). What percentage of the balls in the bag are red?
    A 33.33% Correct Answer Incorrect Answer
    B 44.55% Correct Answer Incorrect Answer
    C 66.67% Correct Answer Incorrect Answer
    D 84.33% Correct Answer Incorrect Answer
    E None of these Correct Answer Incorrect Answer

    Solution

    ATQ,

    Let the number of red balls be 'n'. ATQ, {n/(n+2)} ├Ч {(n-1)/(n+1)} = (2/5) Or, 5(n┬▓ - n) = 2(n┬▓ + 3n + 2) Or, 5n┬▓ - 5n - 2n┬▓ - 6n - 4 = 0 Or, 3n┬▓ - 11n - 4 = 0 Or, 3n┬▓ - 12n + n - 4 = 0 Or, 3n(n - 4) + 1(n - 4) = 0 Or, (3n + 1)(n - 4) = 0 So, n = -1/3 or n = 4 So, n = 4 Total balls = 4 + 2 = 6 Required percent = (4/6) ├Ч 100 = 66.67%

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