Finding probability using conditional probability

Question

Two cards are drawn from a deck of four red cards labeled $1$, $2$, $3$, $4$ and four white cards labeled $1$, $2$, $3$, $4$. Find probability that two cards are white given that at least one card is labeled $2$.

My try: Let $A : \text{two cards are white} , B: \text{At least one card is labeled 2}$ . So the answer is $P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{3}{7+6} = \frac{3}{13}$. Is my answer correct? What are the alternative solutions to this question?

Answer 1

Let $\Omega_0=\{1,2,3,4;1',2',3',4'\}$ be the (set of) cards. (The cards with the prime are red.)

Let $\Omega$ be the set of (ordered) tuples $(a,b)$ with $a,b\in \Omega_0$, $a\ne b$. Let $\Bbb P$ be the uniform probability. (Each subset of $\Omega$ is an event / is measurable.)

Let $W\subset \Omega$ be the set of tuples $(a,b)$ with "white" $a,b\in\{1,2,3,4\}$. Let $T$ be the event of all tuples $(a,b)$ in $\Omega$, so that either $a\in\{2,2'\}$ or $b\in\{2,2'\}$. We can list the elements of $T$, then count them. Or just count them:

• for each $a\in\{2,2'\}$ there are $8-1$ chances for $b$ with $(a,b)\in T$, this leads to a count of $2\cdot 7$ tuples.
• for each $b\in\{2,2'\}$ there are $8-1$ chances for $a$ with $(a,b)\in T$, this leads to a count of $2\cdot 7$ tuples.
• the two tuples $(2,2')$ and $(2',2)$ were counted twice.

So $|T| = 14+14-2 = 26$.

Now let us count the elements in $W\cap T$. Let $(a,b)$ be an element in this intersection. Then either $a=2$ and $b\in \{1,3,4\}$, or $b=2$ and $a\in \{1,3,4\}$. And conversely. (No case is counted twice.) We obtain $3+3=6$ cases.

The wanted conditional probability is: $$ \Bbb P(W|T)=\frac{|W\cap T|}{|T|}=\frac{6}{26}=\frac 3{13}\ . $$

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This is more or less the same solution. (Tuples $(a,b)$ were used to model the situation, instead of sets $\{a,b\}$ as in the OP. Just some more sentences...)