Question
A bag has ((x + 2)) red, ((x + 5)) green and ((x + 8))
yellow balls. Find the probability of taking, without replacement, out 1 red, 2 green and 2 yellow balls if the probability of taking out 1 yellow ball at random from the bag is ((2/5)).Solution
According to question: ((2/5) = (x + 8)C1/(3x + 15)C1) Or, ((2/5) = {(x + 8)/(3x + 15)}) Or, (6x + 30 = 5x + 40) Or, (x = 10) So, number of red balls = ((10 + 2) = 12) Number of green balls = ((10 + 5) = 15) And, number of yellow balls = ((10 + 8) = 18) Required probability = ((12C1 × 15C2 × 18C2)/45C5) = ({(12 × 105 × 153)/1221759}) = (192780/1221759) = (1540/9759)
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