There are 'x' one-rupee coins, 'x + 5' two-rupee coins, and 'x - 5' five-rupee coins in a purse. Two coins are drawn from the purse without replacement, and the probability that neither coin is a two-rupee coin is (66/185). If 15 more ten-rupee coins are added to the purse, after which 2 coins are withdrawn without replacement, what would be the probability that one coin is a one-rupee coin while the other is a ten-rupee coin?

Solution

ATQ, Total number of coins in the purse = x + x + 5 + x - 5 = '3x' Probability that if two coins are withdrawn, neither is two rupee coins = {(2x - 5) /3x} × {(2x - 5 - 1) /(3x - 1) } = (66/185) Or, 185 × {(2x - 5) /3x} × {(2x - 6) /(3x - 1) } = 66 Or, 185 × (2x - 5) (2x - 6) = 66 × 3x × (3x - 1) Or, 185 × (4x²- 12x - 10x + 30) = 198x × (3x - 1) Or, 185 × (2x²- 11x + 15) = 99x × (3x - 1) Or, 370x²- 2035x + 2775 = 297x²- 99x Or, 73x²- 1936x + 2775 = 0 Or, 73x²- 1825x - 111x + 2775 = 0 Or, 73x(x - 25) - 111(x - 25) = 0 Or, (73x - 111) (x - 25) = 0 So, x = 25 or x = (111/73) Since, number of coins has to be an integer, x = 25 So, total number of coins in the purse = 25 × 3 = 75 Number of one rupee coins in the purse = 25 And number of coins in the purse after adding 15 ten rupees coins = 15 + 75 = 90 So, required probability = (25/90) × (15/89) × 2 = (25/267)