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?

$\frac{37}{44}$

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 - \(\frac{7 }{44}\)

= \(\frac{37 }{44}\)