# Problems involving Conditional Probabilities and/or Combination

1. In a poker hand consisting of 5 cards, find the probability of holding 3 aces.

Solution A: $$\frac{^{4}C_3 \times\ ^{48}C_2}{^{52}C_5}$$ Solution B (Does not work): $$\frac{4}{52}\times\frac{3}{51}\times\frac{2}{50}\times\frac{48}{49}\times\frac{47}{48}$$

1. If 3 books are picked at random from a shelf containing 5 novels, 3 books of poems, and a dictionary, what is the probability that: 2 novels and 1 book of poems are selected?

Solution A: $$\frac{^{5}C_2 \times\ ^{3}C_1}{^{9}C_3}$$ Solution B (Does not work): $$\frac{5}{9}\times\frac{4}{8}\times\frac{3}{7}$$

Why do solutions B not give the same answer as solutions A?

## 1 Answer

A probability is the number of favorable cases divided by the number of possible cases (if each case is equally likely). For the first example, the number of hands containing exactly 3 aces is: $$\underbrace{4 \cdot 3 \cdot 2}_\text{nb ways to pick and arrange 3 out of 4 cards} / \underbrace{(3 \cdot 2 \cdot 1)}_\text{nb ways to arrange 3 cards (correction because order doesn't matters)} ... \\ \cdot \underbrace{48 \cdot 47}_\text{nb ways to pick and arrange 2 out of 48 cards} / \underbrace{(2 \cdot 1)}_\text{nb ways to arrange 2 cards (correction because order doesn't matters)} = 4'512$$ While the number of possible 5-cards hands is: $$\underbrace{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48}_\text{nb ways to pick and arrange 5 out of 52 cards} / \underbrace{(5 \cdot 4 \cdot 3 \cdot 2 \cdot 1)}_\text{nb ways to arrange 5 cards (correction because order doesn't matters)} = 25'908'960$$

Dividing these two terms yields solution A. Solution B is wrong, because it implies that the order in which the cards/books are picked matters (simplifying a bit, because if order mattered, your solution would still be slightly off). You can see that to reconcile solution B with solution A, you need to adjust your numerator and denominator to take into account that the order doesn't matter.