you've drawn your two cards and, tragically, neither of them is an ace.
Your friend is feeling generous: she lets you draw ten additional cards. Still no luck. You know there are 40 cards remaining in the deck, of which four are aces.
If your friend now draws two cards, what's the probability she gets exactly one ace?
I thought the prob are independent: so the prob to get an ace on the first card is 4/40, and the second 36/40 (not being an ace), I multiplied them and give me some prob. but I am not sure if the logic I am using is right. Someone does it know if this is the correct logic?
thanks
Best Answer
Suppose the probability that she gets exactly one ace is P.
Then P= probability (that first card is an ace and second one is not) +
probability (that first card is not an ace but second is an ace) =$\frac{4}{40}\frac{36}{39}+\frac{36}{40}\frac{4}{39}$ =$12/65$
That's the answer.