Question

Three cards are drawn successfully, without replacement from a pack of 52 well shuffled cards. What is the probability that the first two cards are kings and the third card drawn is an ace?

Answer 1

For solving this question you should know about the probability concept. In this problem we will first take the probability for a single card that can be king and then if we reduce that then a new number of cards will be there. Then again, we will find the probability for being a king again in the rest of the cards. And then if two kings are drawn, then again, a new number of cards will be there. So, we will find the probability for being the next card ace. And then finally we will multiply these all with each other to find our result.

Complete step-by-step solution:
According to the question from a pack of 52 cards three cards are drawn successfully, without replacement from that pack. Then we have to find the probability that the first two cards are king and the third card drawn is an ace. So,
Total number of cards in a pack = 52.
Let A denote the event that the card drawn is a king, and K is the event that the card drawn is an ace. Now, probability for the first card being a king is:
$P\left( A \right)=\dfrac{\text{number of possible events}}{\text{number of total events}}$
Since, there are 4 king in a pack of cards, so,
$P\left( A \right)=\dfrac{4}{52}$
Now, if one card has been drawn from the pack, there are now 51 total cards. And the total number of events will now be 51. Now, the probability for the second card being a king is:
$P\left( \dfrac{A'}{A} \right)=\dfrac{3}{51}$
Since one king is also drawn from that, so the number of kings are 3 here. Now the total number of cards is 50, because 2 kings have already been drawn from that, so the new number of total cards has been reduced.
Now there are 4 aces in a pack. Now the probability for being an ace is:
$P\left( \dfrac{K}{AA'} \right)=\dfrac{4}{50}$
So, by the probability multiplication law:
$P\left( \dfrac{K}{AA'} \right)=\dfrac{4}{52}\times \dfrac{3}{51}\times \dfrac{4}{50}=\dfrac{2}{5525}$
So, the probability is $\dfrac{2}{5525}$.

Note: While solving this question you have to remember the properties of probabilities and always you have to calculate the new number of total events for every new probability because in most of the cases of cards every event has a new number of total events if any card is drawn from that.