I have this task about cards:
Consider choosing a card from a well-shuffled standard deck of 52
playing cards.
- Suppose that, after the first extraction, the card
is not reinserted in the deck. What is the probability that the second
card is an ace?- Suppose that, after the first extraction, the card
is reinserted in the deck. What is the probability that the second
card is an ace?
For the first case, I've been thinking about $$\left(\frac{4*3}{52*51}\right) + \left(\frac{48*4}{52*51}\right)$$ Where I counted two examples: a)the ace was drawn already, and b) it wasn't. For the second part of the task, would it be $$\left(\frac{4*4}{52*52}\right) + \left(\frac{52*4}{52*52}\right)$$ I'm pretty sure that I'm assuming wrong, but I can't come up with anything else. Any help would be great!
Best Answer
As Ant said in the comments, in both cases, the answer is $\frac{4}{52} = \frac{1}{13}$. Your first equation simplifies to this: \begin{align*} \left(\frac{4\cdot 3}{52 \cdot 51}\right) + \left(\frac{48 \cdot 4}{52 \cdot 51}\right) &= \left(\frac{4}{52}\right)\left( \frac{3}{51} + \frac{48}{51}\right) \\ &= \left(\frac{4}{52}\right)(1) \\ &= \frac{1}{13} \end{align*} But this is actually overcomplicating things; drawing the first card doesn't affect the odds at all, regardless of whether or not it's replaced. We don't know anything about what the first card drawn was, so it doesn't add any information, and thus doesn't affect the probability that the second card is an ace.