A basket contains 4 red, 5 blue and 3 green marbles. If three marbles are picked at random what is the probability that at least one of them is blue?
Answer & explanation
Correct answer: option 2
Total number of marbles = 4 + 5 + 3 = 12
probability of getting atleast one marble out of 3 is blue = 1 - ( Probability that all 3 marbles are not of blue color )
Probability that all 3 marbles are not of blue color = \(\frac{7C3}{12C3}\)
Probability of selecting 3 marble other than blue color = 7C3
= \(\frac{7 × 6 × 5}{3 × 2 × 1}\)
= 35
Probability of selecting 3 marble = 12C3
= \(\frac{12 × 11 × 10}{3 × 2 × 1}\)
= 220
Now,
probability of getting atleast one marble out of 3 is blue = 1 - \(\frac{35 }{220}\)
= 1 - \(\frac{7 }{44}\)
= \(\frac{37 }{44}\)