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 of them is blue?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\frac{37}{44}\)
“At least one blue” is the complement of “no blue at all”, which is easier to count: all three marbles must then come from the 7 marbles that are not blue (4 red + 3 green).
Ways to pick any 3 of the 12 marbles: \(\displaystyle {}^{12}C_3=\frac{12\times11\times10}{3\times2\times1}=220\)
Ways to pick 3 of the 7 non-blue marbles: \(\displaystyle {}^{7}C_3=\frac{7\times6\times5}{3\times2\times1}=35\)
P(no blue) \(\displaystyle =\frac{35}{220}=\frac{7}{44}\)
P(at least one blue) \(\displaystyle =1-\frac{7}{44}=\frac{37}{44}\)
Tip: For “at least one”, work out the chance of none and subtract it from 1. Option (4) \(\frac{7}{44}\) is the chance of none, the step before the last.