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?Solution
ATQ,
Let the number of red balls be 'n'. ATQ, {n/(n+2)} × {(n-1)/(n+1)} = (2/5) Or, 5(n² - n) = 2(n² + 3n + 2) Or, 5n² - 5n - 2n² - 6n - 4 = 0 Or, 3n² - 11n - 4 = 0 Or, 3n² - 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) × 100 = 66.67%
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