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 is blue?

$\frac{37}{44}$

The correct answer is Option (4) → $\frac{37}{44}$

We are asked: Probability of picking at least one blue marble from a basket containing:

• Red = 4
• Blue = 5
• Green = 3

Total marbles = 4 + 5 + 3 = 12

Step 1: Use complement probability

$P(\text{at least one blue}) = 1 - P(\text{no blue})$

No blue → only red or green are chosen: 4 + 3 = 7 marbles

Number of ways to choose 3 marbles from 7 non-blue:

$\begin{pmatrix}7\\3\end{pmatrix} = 35$

Total ways to choose any 3 marbles from 12:

$\begin{pmatrix}12\\3\end{pmatrix} = 220$

$P(\text{no blue}) = \frac{35}{220} = \frac{7}{44}$

Step 2: Probability of at least one blue

$P(\text{at least one blue}) = 1 - \frac{7}{44} = \frac{37}{44}$