StArAnGeL
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?
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ChrisV
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1/((52)(51)(50))
ChrisV
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1/32600
StArAnGeL
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Sorry Thats not the answer
ChrisV
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really?
StArAnGeL
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Yeah !
The answer is 2/5525
But i need to know how these kind of problems are solved
ChrisV
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of well the fact is your chance of the first card being a king is 1/52
ChrisV
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the second card being a king would be 1/51
ChrisV
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and the third card being an ace would be 1/50
ChrisV
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how they came up with that answer i do not know
ChrisV
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unless they are somehow figuring in that there are 4 of each
StArAnGeL
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Oh Ok
there are 4kings So chance of the first card being a king is 4/52
StArAnGeL
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and then 3/51
StArAnGeL
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and ace will be 4/50
StArAnGeL
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So multiply and its 2/5525
gogind
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ah you got it...nvm
StArAnGeL
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But Btw do you know Why we are multiplying ?
ChrisV
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yes
StArAnGeL
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Um Why ?
ChrisV
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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
ChrisV
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occur
ChrisV
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like if you have 10 books and 2 are school books
ChrisV
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whats the probability that the first book you pull out of your bag would be a school book
ChrisV
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2/10=1/5
StArAnGeL
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yes But we didnt multiply anything here ?
ChrisV
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so say you have taken 1 school book out already
ChrisV
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whats the chance of pulling both school book out of the bag in a row
ChrisV
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well we already know the first one was 2/10
ChrisV
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well that leaves 1/.9
ChrisV
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so you multiply 2/10 x 1/9
ChrisV
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the probability of pulling them both out in order would be 2/90 or 1/45
ChrisV
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so out of every 90 times you try it, the probability of succedding is 2 times
StArAnGeL
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OH Ohkay
Thanks a lot =)
FoolForMath
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we are multiplying because these are independent events.
\( P(A\cap B)=P(A) \times P(B) \)
ChrisV
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yes
Directrix
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Disclaimer: I know this problem has been solved. I am just practicing here.
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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.