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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?

Mathematics
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1/((52)(51)(50))
1/32600
Sorry Thats not the answer

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Other answers:

really?
Yeah ! The answer is 2/5525 But i need to know how these kind of problems are solved
of well the fact is your chance of the first card being a king is 1/52
the second card being a king would be 1/51
and the third card being an ace would be 1/50
how they came up with that answer i do not know
unless they are somehow figuring in that there are 4 of each
Oh Ok there are 4kings So chance of the first card being a king is 4/52
and then 3/51
and ace will be 4/50
So multiply and its 2/5525
ah you got it...nvm
But Btw do you know Why we are multiplying ?
yes
Um Why ?
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
occur
like if you have 10 books and 2 are school books
whats the probability that the first book you pull out of your bag would be a school book
2/10=1/5
yes But we didnt multiply anything here ?
so say you have taken 1 school book out already
whats the chance of pulling both school book out of the bag in a row
well we already know the first one was 2/10
well that leaves 1/.9
so you multiply 2/10 x 1/9
the probability of pulling them both out in order would be 2/90 or 1/45
so out of every 90 times you try it, the probability of succedding is 2 times
OH Ohkay Thanks a lot =)
we are multiplying because these are independent events. \( P(A\cap B)=P(A) \times P(B) \)
yes
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 52-count 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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