A basket contains 4 red, 5 blue and 3 green marbles. If three marbles are picked up at random, what is the probability that at least one is blue?

$\frac{37}{44}$

The correct answer is Option (1) → $\frac{37}{44}$

Total marbles = 4 + 5 + 3 = 12

Step 1: Use the complement method

At least one blue
= 1 − (no blue marbles)

Non-blue marbles = Red + Green = 4 + 3 = 7

Step 2: Calculate probabilities

$P(\text{no blue}) = \frac{\begin{pmatrix}7\\3\end{pmatrix}}{\begin{pmatrix}12\\3\end{pmatrix}} = \frac{35}{220}$

$P(\text{at least one blue}) = 1 - \frac{35}{220} = \frac{185}{220} = \frac{37}{44}$