A bridge hand (13 cards) is dealt from a standard 52 card deck. Let A be the event that the hand contains four aces. Let B be the event that the hand contains four kings. Find $P(A \cup B)$.
Attempt: The four aces can be dealt in $_4C_4$ different ways. Thus the 9 remaining cards can be selected out of the remaining 48 cards. Thus, $_{48}C_{9}$ ways. From the 52 cards, a bridge hand can be selected in $_{52}C_{13}$ ways. Thus, $$P(A) = \frac{_4C_4 \cdot~_{48}C_9}{_{52}C_{13}}$$
Then we also need to find P(B) which will be equal to $$\frac{_4C_4 \cdot~ _{48}C_9}{_{52}C_{13}}$$ since it is the same only with kings.
Then we need to find $P(A \cup B) = P(A) + P(B) – P(A \cap B)$.
I don't know if I am on the right track. Please can someone help me?
Thank you.
Best Answer
Yes you're right and we have using the same idea
$$P(A\cap B)=\frac{C_8^8\times C_{5}^{44}}{C_{52}^{13}}$$