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?
Answer & explanation
Correct answer: option 4
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
$P(\text{at least one blue}) = 1 - \frac{7}{44} = \frac{37}{44}$