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?

The basket contains:

Red marbles: 4

Blue marbles: 5

Green marbles: 3

Total marbles: $4 + 5 + 3 = 12$

Three marbles are picked at random.

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"At least one blue" means 1 blue, 2 blues, or 3 blues. Calculating all three cases separately requires multiple calculations.

Using the complement:

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

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If no blue marbles are picked, all three marbles must be red or green.

Non-blue marbles = $4 + 3 = 7$

Total ways to pick 3 marbles from 12:

$C(12,3) = \dfrac{12 \times 11 \times 10}{3 \times 2 \times 1}$

$= \dfrac{1320}{6}$

$= 220$

Ways to pick 3 marbles from 7 non-blue marbles:

$C(7,3) = \dfrac{7 \times 6 \times 5}{3 \times 2 \times 1}$

$= \dfrac{210}{6}$

$= 35$

Therefore:

$P(\text{no blue}) = \dfrac{35}{220}$

$= \dfrac{7}{44}$

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

$= \dfrac{44}{44} - \dfrac{7}{44}$

$= \dfrac{37}{44}$

The answer is $\dfrac{37}{44}$ (Option 4).