A basket contains 4 red, 5 blue and 3 green marbles. If 3 marbles are picked at random, what is the probability that either all are green or all are red?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\frac{1}{44}\)
“All green” and “all red” cannot happen together, so their numbers of ways add; every choice of 3 marbles out of 12 is equally likely.
All possible choices: \({}^{12}C_3=\displaystyle\frac{12\times11\times10}{3\times2\times1}=220\)
All green: \({}^{3}C_3=1\); all red: \({}^{4}C_3=4\)
Probability \(=\displaystyle\frac{1+4}{220}=\frac{5}{220}=\frac{1}{44}\)