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Probability and replacement

  1. Sep 4, 2008 #1
    if 5 cards are randomly selected from a standard deck of 52 cards.

    what is the probability that all 5 cards are red if they are selected without replacement?

    and

    what is the probability that all 5 cards are red if they are selected with replacement?


    need help with this!!
     
  2. jcsd
  3. Sep 4, 2008 #2
    Well the probability that the first one is red is 13/52, given that the first one is red if you dont replace then the probability second one is red is 12/51 (because you have 12 red left out of 51). If you replace, well then again its the same probability of 13/52 getting a second red.

    I don't want to ruin it for you so you can do the 3rd, 4th and 5th cards. When you have them all, just multiply them together to get the probability of getting all 5 red with/without replacement.
     
  4. Sep 7, 2008 #3
    thanks!
     
  5. Sep 9, 2008 #4
    where did 13 come from?
     
  6. Sep 9, 2008 #5

    Dick

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    One fourth of the cards in the deck are red. 52/4=13.
     
  7. Sep 9, 2008 #6
    oh thanks.. i have a question.. u kno when u find P when it says find P(a and b)
    find P(a or b)

    but what is this find P(a/b)
     
  8. Sep 9, 2008 #7

    Dick

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    It's P(a and not(b)). It's the set theory difference.
     
  9. Sep 9, 2008 #8
    ?? im confused.. what is the forumula. the line in between is to multiply or?
     
  10. Sep 9, 2008 #9
    if the event it independent or not, you have to use this formula P(A/B)

    im confused..

    and how can you know if its independent or not?
     
  11. Sep 9, 2008 #10

    Dick

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    I told you. The set A/B means "A intersect not(B)". P(not(B))=1-P(B). It's the complement of B. You can only tell if two events are independent by describing what the events are. If one occurring has no effect on the other occurring then they are independent.
     
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