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) → \(\displaystyle\frac{37}{44}\)
“At least one blue” covers every case except “no blue at all”, so P(at least one) = 1 − P(none), which needs only one count instead of three.
The marbles:
Marbles in all: \(4+5+3=12\)
Marbles that are not blue: \(4+3=7\)
The number of ways (to pick 3):
Ways to pick any 3: \(\displaystyle {}^{12}C_{3}=\frac{12\times11\times10}{3\times2\times1}=220\)
Ways to pick 3 with no blue: \(\displaystyle {}^{7}C_{3}=\frac{7\times6\times5}{3\times2\times1}=35\)
The probability (1 − P(none)):
P(no blue) \(\displaystyle =\frac{35}{220}=\frac{7}{44}\)
P(at least one blue) \(\displaystyle =1-\frac{7}{44}=\frac{37}{44}\)
The value \(\frac{7}{44}\) in Option (3) is the probability that no blue marble is picked, not that at least one is.