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

This Question is Closed

StArAnGeL
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry Thats not the answer

StArAnGeL
 3 years ago
Best ResponseYou've already chosen the best response.0Yeah ! The answer is 2/5525 But i need to know how these kind of problems are solved

ChrisV
 3 years ago
Best ResponseYou've already chosen the best response.1of well the fact is your chance of the first card being a king is 1/52

ChrisV
 3 years ago
Best ResponseYou've already chosen the best response.1the second card being a king would be 1/51

ChrisV
 3 years ago
Best ResponseYou've already chosen the best response.1and the third card being an ace would be 1/50

ChrisV
 3 years ago
Best ResponseYou've already chosen the best response.1how they came up with that answer i do not know

ChrisV
 3 years ago
Best ResponseYou've already chosen the best response.1unless they are somehow figuring in that there are 4 of each

StArAnGeL
 3 years ago
Best ResponseYou've already chosen the best response.0Oh Ok there are 4kings So chance of the first card being a king is 4/52

StArAnGeL
 3 years ago
Best ResponseYou've already chosen the best response.0So multiply and its 2/5525

StArAnGeL
 3 years ago
Best ResponseYou've already chosen the best response.0But Btw do you know Why we are multiplying ?

ChrisV
 3 years ago
Best ResponseYou've already chosen the best response.1well to find the probability of something happening you have to multiply the number of times it could possibly occure then divide it by the total number of possibilities

ChrisV
 3 years ago
Best ResponseYou've already chosen the best response.1like if you have 10 books and 2 are school books

ChrisV
 3 years ago
Best ResponseYou've already chosen the best response.1whats the probability that the first book you pull out of your bag would be a school book

StArAnGeL
 3 years ago
Best ResponseYou've already chosen the best response.0yes But we didnt multiply anything here ?

ChrisV
 3 years ago
Best ResponseYou've already chosen the best response.1so say you have taken 1 school book out already

ChrisV
 3 years ago
Best ResponseYou've already chosen the best response.1whats the chance of pulling both school book out of the bag in a row

ChrisV
 3 years ago
Best ResponseYou've already chosen the best response.1well we already know the first one was 2/10

ChrisV
 3 years ago
Best ResponseYou've already chosen the best response.1so you multiply 2/10 x 1/9

ChrisV
 3 years ago
Best ResponseYou've already chosen the best response.1the probability of pulling them both out in order would be 2/90 or 1/45

ChrisV
 3 years ago
Best ResponseYou've already chosen the best response.1so out of every 90 times you try it, the probability of succedding is 2 times

StArAnGeL
 3 years ago
Best ResponseYou've already chosen the best response.0OH Ohkay Thanks a lot =)

FoolForMath
 3 years ago
Best ResponseYou've already chosen the best response.1we are multiplying because these are independent events. \( P(A\cap B)=P(A) \times P(B) \)

Directrix
 3 years ago
Best ResponseYou've already chosen the best response.2Disclaimer: I know this problem has been solved. I am just practicing here.  4 kings, want 2 4 aces, want 1 no replacement 52 cards in deck P(K) = 4/52. The 52count drops to 51 with no replacement. King count drops to 3 P(2ndK) = 3/51. The 51 card count drops to 50. King count drops to 2. Ace count remains at 4. P(Ace) = 4 / 50 Using the Multiplication Principle, the probability of these 3 events happening is (4/52) (3/51) (4/ 50) = 2/5525 . Multiplication Principle states: If an event occurs in m ways and another event occurs independently in n ways, then the two events can occur in m × n ways.
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