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Thread: Probability

  1. #1 Probability 
    Forum Ph.D. Heinsbergrelatz's Avatar
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    can you please help me out with this question please? this is not my homework, and no im not being lazy, just want to learn how to do it, for revision purposes. thank you

    a pack of 16 cards contains 4 cards of each of the colors red,blue,green, and yellow.the four cards of each color are numbered 1,2,3 and 4.

    i)the pack is shuffled and is drawn at random, find as a fraction that it is a red card numbered 4 or a blue card.

    ii)The card is replaced, the pack is shuffled again and two cards are drawn at random. Find as a fraction the probability that
    a)they are both yellow
    b)at least one is blue.

    for number I), i was able to solve it by going (1/4*1/4 + 1/4)=5/16
    BUT its the ii) that im stuck.

    Appreciate the help.


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  3. #2  
    Forum Ph.D.
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    iia) First card is yellow, p=1/4. Second card is yellow (assuming first is) p=3/15.
    Net result = 3/60.

    iib) First card is not blue, p=3/4. Second card is not blue (assuming first not) p=4/5.
    Net result prob(no blue)=3/5.
    Prob(at least one blue)=2/5.


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  4. #3  
    Forum Ph.D. Heinsbergrelatz's Avatar
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    thank you for your help. but are you sure these are positively correct?
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  5. #4  
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    Done both for you, I get same result as other guy, kind of explained it as I have gone through.

    For number one its 1/16 + 4/16, Can be either so you work out both and add.

    P(First Card is Yellow)=1/4
    P(Second card is yellow assuming first was yellow=3/15

    As they both have to be correct you multiply them together.

    P(Both Cards are Yellow)=1/4*3/15=3/60=1/20

    P(First Card is not blue) =12/16
    P(Second Card is not blue, assuming first isn't blue) =12/15

    P(Neither Card is Blue) =12/16*12/15=144/240=3/5

    P(So, at least one card being blue)=2/5
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