anonymous
  • anonymous
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
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
1/((52)(51)(50))
anonymous
  • anonymous
1/32600
anonymous
  • anonymous
Sorry Thats not the answer

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