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?

- anonymous

- schrodinger

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- anonymous

1/((52)(51)(50))

- anonymous

1/32600

- anonymous

Sorry Thats not the answer

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## More answers

- anonymous

really?

- anonymous

Yeah !
The answer is 2/5525
But i need to know how these kind of problems are solved

- anonymous

of well the fact is your chance of the first card being a king is 1/52

- anonymous

the second card being a king would be 1/51

- anonymous

and the third card being an ace would be 1/50

- anonymous

how they came up with that answer i do not know

- anonymous

unless they are somehow figuring in that there are 4 of each

- anonymous

Oh Ok
there are 4kings So chance of the first card being a king is 4/52

- anonymous

and then 3/51

- anonymous

and ace will be 4/50

- anonymous

So multiply and its 2/5525

- anonymous

ah you got it...nvm

- anonymous

But Btw do you know Why we are multiplying ?

- anonymous

yes

- anonymous

Um Why ?

- anonymous

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

- anonymous

occur

- anonymous

like if you have 10 books and 2 are school books

- anonymous

whats the probability that the first book you pull out of your bag would be a school book

- anonymous

2/10=1/5

- anonymous

yes But we didnt multiply anything here ?

- anonymous

so say you have taken 1 school book out already

- anonymous

whats the chance of pulling both school book out of the bag in a row

- anonymous

well we already know the first one was 2/10

- anonymous

well that leaves 1/.9

- anonymous

so you multiply 2/10 x 1/9

- anonymous

the probability of pulling them both out in order would be 2/90 or 1/45

- anonymous

so out of every 90 times you try it, the probability of succedding is 2 times

- anonymous

OH Ohkay
Thanks a lot =)

- anonymous

we are multiplying because these are independent events.
\( P(A\cap B)=P(A) \times P(B) \)

- anonymous

yes

- Directrix

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.

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