Question

Drawing Three Cards. From a deck of 52 cards, 3 are drawn at random without replacement. What is the probability that 2 are aces and 1 is a king? find answer as 6\5525

Answer 1

Answer:

,There are 4 king and 4 Queen in a deck of 52 playing cards.

2 Kings out of the 4. It can be chosen in 4C2 ways.

4C2 = 4!/2!*2! = 6

1 Queen out of the 4. It can be done in 4C1 ways.

4C1 = 4!/3!*1! = 4

3 cards are to be chosen from a deck of 52 cards. It can be done in 52C3 ways.

52C3 = 52!/3!*49! = 22100

Probability = 4C2 * 4C1 / 52C3 = 6*4/22100 = 6/5525 or (0.001085972850)

Answer 2

$</p><p></p><p></p><p></p><p></p><p>Join / Login</p><p></p><p>Question</p><p></p><p>Three cards are drawn successively, without replacement from a pack of 52well-shuffled cards. What is the probability that the first two cards are kings and the third card drawn is an ace?</p><p></p><p>Easy</p><p></p><p>Open in App</p><p></p><p>Solution</p><p></p><p></p><p></p><p>Verified by Toppr</p><p></p><p>Let K denote the event that the card drawn is king and A be the event that the card drawn is an ace.</p><p></p><p>Now, </p><p></p><p>P(K)=524</p><p></p><p>Also, P(K I K) is the probability of second king with the condition that one king has already been drawn.</p><p></p><p>Now there are three kings in (52−1)=51cards. </p><p></p><p>P(K∣K)=513</p><p></p><p>Again, P(A∣KK) is the probability of third drawn card to be an ace, with the condition that two kings have already been drawn. Now, there are four aces in remaining 50cards. P(A∣KK)=504 By multiplication law of probability, we have P(KKA)=P(K)P(K∣K)P(A∣KK)=524×513×504=55252</p><p></p><p>$