I'm working on this question which has got me stumped. Struggling to workout the probabilities for each part of the conditional equation. Here's the question (from Applied Finite Mathematics if anyone's interested)
"Three cards are randomly drawn without replacement from a standard deck of 52 cards.
What is the probability of drawing an ace on the third draw given that at least one ace was drawn on the first two draws"
The answer is given as 49/825 or 0.059
How this is arrived at? In particular how to derive the probabilities for each part of the equation. Thank you.
Let $X$ be the number of aces drawn in the first two draws. Now let us define the following events: $$A = \{ \text{you draw an ace in the third draw}\}$$ $$B = \{X = 1\}$$ $$C = \{X = 2\}$$
You want to know $\mathbb{P}(A | \, X \geq 1)$ which can be written as $$\mathbb{P}(A | \, B \cup C) = \frac{\mathbb{P}(A \cap (B \cup C))}{\mathbb{P}(B \cup C)} = \frac{\mathbb{P}(A \cap B) + \mathbb{P}(A \cap C)}{\mathbb{P}(B) + \mathbb{P}( C)} = \frac{\mathbb{P}(A | \, B)\mathbb{P}(B) + \mathbb{P}(A |\, C)\mathbb{P}(C)}{\mathbb{P}(B) + \mathbb{P}( C)}$$
Now it suffices to calculate each element: $$\mathbb{P}(A |\, B)=\frac{3}{50}$$ $$\mathbb{P}(A |\, C)=\frac{2}{50}$$ $$\mathbb{P}(B) = \frac{\binom{4}{1}\binom{48}{1}}{\binom{52}{2}}=\frac{2 \cdot 4 \cdot 48}{52 \cdot 51}$$ $$\mathbb{P}(C) = \frac{\binom{4}{2}\binom{48}{0}}{\binom{52}{2}}=\frac{4 \cdot 3}{52 \cdot 51}$$