Question
A poly-bag contains (3a - 4) red, (2a - 3) white, and (a
- 1) black rubber bands. Two rubber bands are drawn randomly from the bag, and the probability that both are red is 7/30. Based on this information, determine the probability that at most one of the two drawn rubber bands is white.Solution
ATQ,
Number of rubber bands in the Poly-bag = 3x - 4 + 2x - 3 + x - 1 = 6x - 8
ATQ:
{(3a-4)/(6a-8) × (3a-5)/(6a-9)} = 7/30
Or, 30 X (9a² - 27a + 20) = 7 X (36a² - 102a + 72)
Or, 270a² - 810a + 600 = 252a² - 714a + 504
Or, 18a² - 96a + 96 = 0
Or, 3a² - 16a + 16 = 0
Or, (3a - 4) (a - 4) = 0
So, 'a' = (4/3) or 'a' = 4
But 'a' should be an integer. So, 'a' = 4
Number of white rubber bands = 2 X 4 - 3 = 5
Total number of rubber bands = 6 X 4 - 8 = 16
Required probability:
Probability that none of the rubber bands is white + Probability that one of the rubber bands is white
= (11/16) X (10/15) + 2 X (5/16) X (11/15)
= (110 + 110) ÷ 240
= (220/240) = (11/12)
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