Skip to main contentSkip to solution

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

Total caps=2+4+5+1=12\text{Total caps}=2+4+5+1=12

Using the combination formula nCr=n!r!(n−r)!{}^{n}C_{r}=\dfrac{n!}{r!(n-r)!}, the number of ways to pick any 3 caps is

12C3=12×11×103×2×1=220{}^{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:

4C2×5C1=6×5=30{}^{4}C_{2}\times{}^{5}C_{1}=6\times5=30


Probability=favourabletotal=30220=322\text{Probability}=\dfrac{\text{favourable}}{\text{total}}=\dfrac{30}{220}=\dfrac{3}{22}

Keyboard Shortcuts

  • Left arrow: Previous question
  • Right arrow: Next question
  • S key: Jump to solution
  • Q key: Jump to question