A hand $H$ of 5 cards is chosen randomly from a standard deck of 52. Let $E_1$ be the event that H has at least one King and let $E_2$ be the event that $H$ has at least 2 Kings. What is the conditional probability $\mathbb P(E_2|E_1)$?
Solution:
$$\mathbb P(E_2|E_1)=1-\frac{4\binom{48}4}{\binom{52}5-\binom{48}5}.$$
Would this be a correct solution (obtained by looking at a similar problem)? I'm a bit confused by this solution, can someone walk me through it so I can understand.
Best Answer
We have by the definition of conditional probability $$\Pr(E_2|E_1)=\frac{\Pr(E_1\cap E_2)}{\Pr(E_1)}.$$ We calculate the two probabilities on the right-hand side.
The probability of $E_1$ is $1$ minus the probability of no Kings. You calculated $\Pr(E_1)$ correctly.
For the probability of $E_1\cap E_2$, which is just the probability of $E_2$, use the same basic strategy. The number of $1$ King hands is $\binom{4}{1}\binom{48}{4}$, for we have to choose $4$ non-Kings to go with the King. So the probability of at least $2$ Kings is $$1-\frac{\binom{4}{0}\binom{48}{5}+\binom{4}{1}\binom{48}{4}}{\binom{52}{5}}.$$