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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?

Solution

✅ Correct Option: 1

The basket contains:

  • Red marbles: 4
  • Blue marbles: 5
  • Green marbles: 3

Total marbles = 4 + 5 + 3 = 12


For "at least one blue" problems, use the complement:

P(at least one blue) = 1 - P(no blue marbles)


Total ways to pick 3 marbles from 12:

C(12,3)=12×11×103×2×1C(12,3) = \dfrac{12 \times 11 \times 10}{3 \times 2 \times 1}

=13206= \dfrac{1320}{6}

=220= 220


Ways to pick 3 marbles with no blue (only red and green):

Non-blue marbles = 4 + 3 = 7

C(7,3)=7×6×53×2×1C(7,3) = \dfrac{7 \times 6 \times 5}{3 \times 2 \times 1}

=2106= \dfrac{210}{6}

=35= 35


P(no blue marbles) = 35220=744\dfrac{35}{220} = \dfrac{7}{44}


P(at least one blue) = 1−7441 - \dfrac{7}{44}

=4444−744= \dfrac{44}{44} - \dfrac{7}{44}

=3744= \dfrac{37}{44}

Therefore, the probability that at least one marble is blue is 3744\dfrac{37}{44}.

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