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    Well, good news is that I just have to do this module and another module next year and then so long probability and stats.

    Can somebody help me with this

    1. Two cards are drawn, without replacement, one at a time from a pack of 53 cards.
    a) what are the probability that they are both clubs.

    I'm guessing its this
    P(c|c) = \frac{P(c \cup c)}{P(c)} So

    P(c|c) \times P(c) = P(c \cap c)

    Then just work out P(c|c) and P(c)?

    I don't really know why this works or if it works, but hey thats stats.

    b) what is the probability the second card is a club,

    I'm thinking something similar but with P(c|c^c) has to come in somehow, dammit.

    P.S. Well, it looks like I should focus on rote learning i.e. do tons of example so I can do rules I don't understand.
    P.P.S. Although, the proofs are easy to follow its just the application of the stuff.
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    It might be more helpful, rather than writing 'c', to write what you actually want:

     \mathbf{P} (\mathrm{Second \ card \ is \ a \ club} | \mathrm{First \ card \ is \ a \ club}) = \dfrac{\mathbf{P} (\mathrm{First \ card \ and \ second \ card \ are \ clubs})}{\mathbf{P} (\mathrm{First \ card \ is \ a \ club})}

    Alternatively, a tree diagram could help with this problem, if you're finding it tough.
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    (Original post by Simplicity)
    Well, good news is that I just have to do this module and another module next year and then so long probability and stats.

    Can somebody help me with this

    1. Two cards are drawn, without replacement, one at a time from a pack of 53 cards.
    a) what are the probability that they are both clubs.

    I'm guessing its this
    P(c|c) = \frac{P(c \cup c)}{P(c)} So

    P(c|c) \times P(c) = P(c \cap c)

    Then just work out P(c|c) and P(c)?

    I don't really know why this works or if it works, but hey thats stats.

    b) what is the probability the second card is a club,

    I'm thinking something similar but with P(c|c^c) has to come in somehow, dammit.

    P.S. Well, it looks like I should focus on rote learning i.e. do tons of example so I can do rules I don't understand.
    P.P.S. Although, the proofs are easy to follow its just the application of the stuff.
    P(c|c)

    why are you using conditional probability? this is far simpler.
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    :lolwut: is this really first year University level? It's barely much harder than the stats in GCSE Maths?
    Anyway, just draw a tree diagram showing the possible suits that the card drawn could be for each of the times a card is drawn, work out all the different possibilities and the probabilities of each of these, and then your answers should be easy to find from there.
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    (Original post by GHOSH-5)
    It might be more helpful, rather than writing 'c', to write what you actually want:

     \mathbf{P} (\mathrm{Second \ card \ is \ a \ club} | \mathrm{First \ card \ is \ a \ club}) = \dfrac{\mathbf{P} (\mathrm{First \ card \ and \ second \ card \ are \ clubs})}{\mathbf{P} (\mathrm{First \ card \ is \ a \ club})}

    Alternatively, a tree diagram could help with this problem, if you're finding it tough.
    No not a tree diagram. I'm lazy, but yeah thats what it meant so is that correct.

    What about the other one? do you work at P(club| first not club) and just add it to it?
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    (Original post by Revolution is my Name)
    :lolwut: is this really first year University level? It's barely much harder than the stats in GCSE Maths?
    Anyway, just draw a tree diagram showing the possible suits that the card drawn could be for each of the times a card is drawn, work out all the different possibilities and the probabilities of each of these, and then your answers should be easy to find from there.
    Not a tree, the lecturer did it using conditional probability but sadly I didn't copy it because I thought I would be able to reproduce it later. Which, didn't happen. No, this is just one problem, to be fair I could do it with a tree diagram like when I did S2, but that would mean drawing a tree diagram and stuff.
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    (Original post by Simplicity)
    No not a tree diagram. I'm lazy, but yeah thats what it meant so is that correct.

    What about the other one? do you work at P(club| first not club) and just add it to it?
    Sort of:

     \mathbf{P}(\mathrm{2nd} = \mathrm{c}) = \mathbf{P}(\mathrm{2nd} = \mathrm{c} \cap \mathrm{1st} = \mathrm{c}) + \mathbf{P}(\mathrm{2nd} = \mathrm{c} \cap \mathrm{1st} \not= \mathrm{c})

     = \mathbf{P}(\mathrm{2nd} = \mathrm{c}|\mathrm{1st} = \mathrm{c})\mathbf{P}(\mathrm{1s  t} = \mathrm{c}) + \mathbf{P}(\mathrm{2nd} = \mathrm{c}|\mathrm{1st} \not= \mathrm{c}) \mathbf{P}(\mathrm{1st} \not= \mathrm{c})

    Seems a bit pointless using conditional probabilities though.
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    For the first part, I guess you'd just do \dfrac{13}{53} \times \dfrac{12}{52}

    For the second part, I'd add the probability of both being clubs and the probability of only the second one being a club. i.e. \dfrac{13}{53} \times  \dfrac{12}{52} + \dfrac{40}{53} \times \dfrac{13}{52}

    That is ofcourse, assuming there are 53 cards in a pack.
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    (Original post by GHOSH-5)
    Sort of:

     P(\mathrm{2nd} = \mathrm{c}) = P(\mathrm{2nd} = \mathrm{c} \cap \mathrm{1st} = \mathrm{c}) + P(\mathrm{2nd} = \mathrm{c} \cap \mathrm{1st} \not= \mathrm{c})

     = P(\mathrm{2nd} = \mathrm{c}|\mathrm{1st} = \mathrm{c})P(\mathrm{1st} = \mathrm{c}) + P(\mathrm{2nd} = \mathrm{c}|\mathrm{1st} \not= \mathrm{c}) P(\mathrm{1st} \not= \mathrm{c})
    Thanks, as I thought that was how you do it.

    I will probably post later with other problems. So watch this space

    P.S. I probably shouldn't bunk the problem classes.
 
 
 
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