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StArAnGeL
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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?
 2 years ago
 2 years ago
StArAnGeL Group Title
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?
 2 years ago
 2 years ago

This Question is Closed

ChrisV Group TitleBest ResponseYou've already chosen the best response.1
1/((52)(51)(50))
 2 years ago

StArAnGeL Group TitleBest ResponseYou've already chosen the best response.0
Sorry Thats not the answer
 2 years ago

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

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

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

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

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

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

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

StArAnGeL Group TitleBest ResponseYou've already chosen the best response.0
and then 3/51
 2 years ago

StArAnGeL Group TitleBest ResponseYou've already chosen the best response.0
and ace will be 4/50
 2 years ago

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

gogind Group TitleBest ResponseYou've already chosen the best response.0
ah you got it...nvm
 2 years ago

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

ChrisV Group TitleBest ResponseYou've already chosen the best response.1
well 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
 2 years ago

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

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

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

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

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

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

ChrisV Group TitleBest ResponseYou've already chosen the best response.1
well that leaves 1/.9
 2 years ago

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

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

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

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

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

Directrix Group TitleBest ResponseYou've already chosen the best response.2
Disclaimer: 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.
 2 years ago
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