A box contains 2 blue caps, 4 red caps, 5 green caps and 1 yellow cap. If three caps are picked at random without replacement, what is the probability that two are red and one is green?
Solution
✅ Correct Option: 4
$\text{Total caps}=2+4+5+1=12$ Using the combination formula ${}^{n}C_{r}=\dfrac{n!}{r!(n-r)!}$, the number of ways to pick any 3 caps is ${}^{12}C_{3}=\dfrac{12\times11\times10}{3\times2\times1}=220$ We want 2 red out of the 4 red, and 1 green out of the 5 green: ${}^{4}C_{2}\times{}^{5}C_{1}=6\times5=30$ $\text{Probability}=\dfrac{\text{favourable}}{\text{total}}=\dfrac{30}{220}=\dfrac{3}{22}$