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    * A and B are mutually exclusive and c is independant of both A and B, given that: P(A) = 0.4, P(B) = 0.3 and P(C) = 0.2, calculate P(A \cup B \cup C)

    i used the given formula: P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A n B) - P(A n C) - P(B n C) + P(A n B n C)

    to get 0.4+0.3+0.2-0-0-0=0.9 is this correct? i got feeling its not

    * f(x)=3/4(2x-x^2) 0 less than or equal to x less than or equal 2

    determine E[3X+4]:

    i get 3E[x]+4
    = \int x\frac{3}{4}(2x-x^2) + \int 4\frac{3}{4}(2x-x^2)dx|
    =3

    * in a game show a player has to answer 3 questions right, the prob he gets first question right is 0.6 and if he gets a question right the prob he gets next quest right increase by 0.1, if he gets question wrong prob he gets question right decrease by0.1, the contestant must get atleast 2 questions correct to progress to next aruund, whats prob he progresses?

    i used a tree diagram prob the wrong method by i get 63/125 is this correct...

    if there are 3 contestants taking part in the quiz, suppose each contestant performance is indep of the other contestants and is given by the above question? whats prob only 2 progress ?

    i have no idea on this one but got feeling it something like 63/125*63/125*62/125 but think thats wrong
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    (Original post by Amiro)
    i used the given formula: P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cup B) - P(A \cup C) - P(B \cup C) + P(A n B n C)
    Isn't it
    P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + 2P(A \cap B \cap C) ?

    Draw a venn diagram to try to visualise it. I'm likely wrong though
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    (Original post by nota bene)
    Isn't it
    P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + 2P(A \cap B \cap C) ?

    Draw a venn diagram to try to visualise it. I'm likely wrong though
    now the formula is given and proved but it says in the questions independant and mutually exclusive so am confused now
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    (Original post by Amiro)
    * A and B are mutually exclusive and c is independant of both A and B, given that: P(A) = 0.4, P(B) = 0.3 and P(C) = 0.2, calculate P(A \cup B \cup C)

    i used the given formula: P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cup B) - P(A \cup C) - P(B \cup C) + P(A n B n C)

    to get 0.4+0.3+0.2-0-0-0=0.9 is this correct? i got feeling its not
    No it's wrong. Please explain how these terms are supposed to correspond to those in the formula. (There aren't even the same number of terms, so something ain't right...)
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    (Original post by DFranklin)
    No it's wrong. Please explain how these terms are supposed to correspond to those in the formula. (There aren't even the same number of terms, so something ain't right...)
    sorry original formual was wrong i met 'n' not 'u'
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    (Original post by Amiro)
    * f(x)=3/4(2x-x^2) 0 less than or equal to x less than or equal 2

    determine E[3X+4]:

    i get 3E[x]+4
    = \int x\frac{3}{4}(2x-x^2) + \int 4\frac{3}{4}(2x-x^2)dx|
    =3
    If 0<=X<=2, 3X+4 must be at least 4, so E[3X+4] must also be at least 4. (You've shown no working so I can't see where you've gone wrong).

    * in a game show a player has to answer 3 questions right, the prob he gets first question right is 0.6 and if he gets a question right the prob he gets next quest right increase by 0.1, if he gets question wrong prob he gets question right decrease by0.1, the contestant must get atleast 2 questions correct to progress to next aruund, whats prob he progresses?

    i used a tree diagram prob the wrong method by i get 63/125 is this correct...
    I don't think so. Again, you've shown no working, so...

    if there are 3 contestants taking part in the quiz, suppose each contestant performance is indep of the other contestants and is given by the above question? whats prob only 2 progress ?

    i have no idea on this one but got feeling it something like 63/125*63/125*62/125 but think thats wrong
    Yes, that's wrong. Why did you think it might be right? (Where did 62/125 come from?).
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    (Original post by Amiro)
    sorry original formual was wrong i met 'n' not 'u'
    To repeat myself:

    Your answer is wrong. Please explain how these terms are supposed to correspond to those in the formula. There aren't even the same number of terms.

    To spell it out: you wrote 0.4+0.3+0.2-0-0-0=0.9. Which of these terms corresponds to P(A n B)? Which corresponds to P(A n C)? etc?

    It is very difficult to give meaningful help to someone who doesn't post any of their working.
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    (Original post by DFranklin)
    If 0<=X<=2, 3X+4 must be at least 4, so E[3X+4] must also be at least 4. (You've shown no working so I can't see where you've gone wrong).

    I don't think so. Again, you've shown no working, so...

    Yes, that's wrong. Why did you think it might be right? (Where did 62/125 come from?).
    lol sowee i try just hard using latex
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    (Original post by DFranklin)
    If 0<=X<=2, 3X+4 must be at least 4, so E[3X+4] must also be at least 4. (You've shown no working so I can't see where you've gone wrong).

    I don't think so. Again, you've shown no working, so...

    Yes, that's wrong. Why did you think it might be right? (Where did 62/125 come from?).
    integrated frmo 2 and 0:
    integrate 3x/4(2x-x^2) + 4*3/4(2x-x^2)dx which gave me 3 but i think i shud not have integrated the last part and just have plusd 4 on:confused:
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    (Original post by DFranklin)
    I don't think so. Again, you've shown no working, so...
    i done a tree diagram from right(0.6) wrong(0.4) then carry on and multippled the branches with 2 or or rights and added them together
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    (Original post by Amiro)
    integrated frmo 2 and 0:
    integrate 3x/4(2x-x^2) + 4*3/4(2x-x^2)dx which gave me 3 but i think i shud not have integrated the last part and just have plusd 4 on:confused:
    You should get the same answer either way. I guess you got the integral wrong.
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    (Original post by Amiro)
    i done a tree diagram from right(0.6) wrong(0.4) then carry on and multippled the branches with 2 or or rights and added them together
    Well, I guess you did it wrong then. Hard to say any more if you won't post working. (Are you seeing a theme here?)
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    (Original post by DFranklin)
    Well, I guess you did it wrong then. Hard to say any more if you won't post working. (Are you seeing a theme here?)
    right(0.6)-right(0.7)-right(0.8)
    -wrong(0.2)
    -wrong(0.3)-right(0.2)
    -wrong(0.8)

    wrong(0.4)-right(0.3)-right(0.4)
    -wrong(0.6)
    -wrong(0.7)-right(0.6)
    -wrong(0.4)

    so atleast 2 rights (0.6*0.7*0.8)+(0.6*0.7*0.2)+(0.6 *0.3*0.2)+(0.4*0.3*0.4)= 63/125
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    (Original post by Amiro)
    Code:
    right(0.6)-right(0.7)-right(0.8)
                               -wrong(0.2)
                -wrong(0.3)-right(0.2)
                                -wrong(0.8)
     
    wrong(0.4)-right(0.3)-right(0.4)
                                 -wrong(0.6)
                  -wrong(0.7)-right(0.6)
                                  -wrong(0.4)
    Next time, use the [code][/code] tags to preserve the formatting for the tree. Previewing your posts is always a good thing too.

    In the case "right-wrong-right", you've calculated the probability of getting the 3rd question right as 0.1 less than the probability of getting the 2nd question wrong. My interpretation of the question is that you want 0.1 less than the probability of getting the 2nd question right.

    (I haven't checked for other mistakes).
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    (Original post by DFranklin)
    Next time, use the [code][/code] tags to preserve the formatting for the tree. Previewing your posts is always a good thing too.

    In the case "right-wrong-right", you've calculated the probability of getting the 3rd question right as 0.1 less than the probability of getting the 2nd question wrong. My interpretation of the question is that you want 0.1 less than the probability of getting the 2nd question right.

    (I haven't checked for other mistakes).
    hmm it says 'if he gets a question wrong the prob he gets the next question right decreases by 0.1 am confused i get what you mean but still confused lol
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    (Original post by Amiro)
    hmm it says 'if he gets a question wrong the prob he gets the next question right decreases by 0.1 am confused i get what you mean but still confused lol
    So, at the start, p(right) = 0.6.

    He gets the first question right. Now p(2nd q right) = 0.6 + 0.1 = 0.7.
    He gets the 2nd question wrong. Now p(3rd q right) = 0.7 - 0.1 = 0.6.
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    (Original post by DFranklin)
    So, at the start, p(right) = 0.6.

    He gets the first question right. Now p(2nd q right) = 0.6 + 0.1 = 0.7.
    He gets the 2nd question wrong. Now p(3rd q right) = 0.7 - 0.1 = 0.6.
    i see i see is that how you do it:eek:
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    (Original post by Amiro)
    i see i see is that how you do it:eek:
    in which case i get 63/100
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    (Original post by Amiro)
    in which case i get 63/100
    Not what I got.
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    (Original post by DFranklin)
    Not what I got.
    81/125 i got this time done it neater
 
 
 
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