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?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{37}{44}\)
“At least one blue” covers one, two or three blue marbles. Its opposite is the single case “no blue marble at all”, so P(at least one blue) \(=1-\) P(no blue).
Counting the marbles (all of them, and those that are not blue):
Total marbles \(=4+5+3=12\)
Marbles that are not blue \(=4+3=7\)
Choosing 3 marbles (any 3, and 3 with no blue):
Ways to pick any 3 of the 12: \(\displaystyle {}^{12}C_{3}=\frac{12\times11\times10}{3\times2\times1}=220\)
Ways to pick 3 with no blue (all from the 7): \(\displaystyle {}^{7}C_{3}=\frac{7\times6\times5}{3\times2\times1}=35\)
The probabilities (P(at least one blue) \(=1-\) P(no blue)):
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 (2) is the probability of getting no blue marble, the opposite case.