# Probability of drawing red cards

A standard deck of cards consists of 26 red cards (hearts and diamonds) and 26 black cards (clubs and spades). Suppose you shuffle such a deck and draw three cards at random without replacement. Let $A_i =$ the event that the $i$th card is a red card, for $i = 1, 2, 3$. Mark each of the following statements as TRUE or FALSE.

(a) $P(A_2)> P(A_1)$

(b) $A_1$ and $A_3$ are independent.

(c) $P(A_1|A_3)< P(A_1)$

I'm having a lot of trouble with part a. Obviously $P(A_1)$ is $26/52$, but $P(A_2)$ is either ($26/51$) or ($25/51$) depending on what happened in $A_1$.

• Actually, $P(A_2) = 26/52$. For an intuitive answer, look at my related answer here. Also, are you looking for answers or hints for b) and c)? – Em. Oct 2 '16 at 3:43
• Formally, $P(A_2) = P(A_1 \cap A_2) + P(A_1^c \cap A_2) = \cdots.$ – BruceET Oct 2 '16 at 3:43
• @Max ohhh, now I see, we're going to be drawing the three cards first, and THEN looking at them all at the same time. I was assuming that we would be drawing them one at a time. – DERPYPENGUIN Oct 2 '16 at 3:46
• And yes, I would appreciate answers or hints for b and c as well. I thought they were straightforward, but I see now that my logic was flawed for the entire problem. – DERPYPENGUIN Oct 2 '16 at 3:48
• @DERPYPENGUIN: No, look at the answer below. Intuitively they can't be, if $A_3$ happens, this affects the probability of $A_1$ having happened since there are fewer red than black cards. The formal demonstration is below. – copper.hat Oct 2 '16 at 5:40

The formal answer is that the first card is red or black with probability $\frac 12$. The probability the second card is red is then $\frac{25}{51} \cdot \frac 12$ (first card is red) + $\frac{26}{51} \cdot \frac 12$ (first card is black)$=\frac 12$, so it is false. The intuitive answer is to draw the two cards. Clearly $P(A_1)=\frac 12$. Before you look at them, swap them. Now $P(A_2)=\frac 12$. Done.
Each permutation of the deck is equally likely. When you select the three cards, each one is equally likely to be red or black from which the probablity follows (${1 \over 2}$).
Hence $pA_1 p A_3 = {1 \over 4}$, however $p (A_1 \text{ and } A_3) = { \binom{26}{2}\over \binom{52}{2}} = {25 \cdot 26 \over 51 \cdot 52 } = {25 \over 102} < {1 \over 4}$.
Intuitively, if you know that $A_3$ occurred, there are fewer red cards than black cards remaining, hence the chance of $A_1$ being red is lessened slightly. It is easy to compute $p [A_1 | A_3] = { p (A_1 \text{ and } A_3) \over p A_3} = {50 \over 102} < {1 \over 2}$