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URGENT! Need 1996+ marking schemes for Maths, P2 Q9

  • 11-06-2011 11:03AM
    #1
    krisityfer
    Closed Accounts Posts: 21 ✭


    Well, after honours Paper 1's... unorthadox questions, I'm going all the way back to 1996 in my exam papers, so I'm looking for marking schemes for Question 9 (Further probability and statistics).

    I found this http://www.studentxpress.ie/papers.htm but it doesn't cover the probability option question, so I'm sort of stuck. Any speedy help would be greatly appreciated, what with the exam being on Monday and all.


Comments

  • #2
    Ficheall
    Registered Users, Registered Users 2, Paid Member Posts: 9,067 ✭✭✭Ficheall


    Can't find the paper online - could you post the question?


  • #3
    eoins23456
    Registered Users, Registered Users 2 Posts: 603 ✭✭✭eoins23456


    www.examinations.ie will have the marking schemes back to 2001 for all the option questions


  • #4
    krisityfer
    Closed Accounts Posts: 21 ✭krisityfer


    eoins23456, I'm aware of that, but it's a '96 question I'm stuck on, and for some reasons SEC won't go back that far.

    Anyway, here is the question:
    In a game of chess against a particular opponent, the probability that Sean wins is 3/5. He plays six games against this opponent. What is the proabability that Sean will: lose the second and fourth games and win the others?

    Now the EDCO papers give the answer as {(3/5)^4}x{(2/5)^2}, which gives 324/15625, and seems fairly intuitive.
    However, I would consider that that only gives the probability that he loses ANY two games and wins ANY four. But, out of the possible combinations of two games which he can lose (6C2, or 15), only one of them involves losing both games 2 and 4. So shouldn't the answer of 324/15625 be multiplied by 1/15?
    So now I'm confused as to which answer is correct, mine or EDCO's. Hence, I need the marking scheme to check, but any other opinions would be greatly appreciated.


  • #5
    Ficheall
    Registered Users, Registered Users 2, Paid Member Posts: 9,067 ✭✭✭Ficheall


    You're overcomplicating it.
    krisityfer wrote: »
    Anyway, here is the question:
    In a game of chess against a particular opponent, the probability that Sean wins is 3/5. He plays six games against this opponent. What is the proabability that Sean will: lose the second and fourth games and win the others?

    So it's WIN LOSE WIN LOSE WIN WIN.
    K, all events are independent. View them independently:

    Probability of winning the first game is 3/5
    probability that he loses the second is 2/5
    probability that he wins the third is 3/5
    probability that he loses the fourth is 2/5
    probability he wins the fifth is 3/5
    probability that he wins the sixth is 3/5

    Probability of all these happening =
    3/5 x 2/5 x 3/5 x 2/5 x 3/5 x 3/5 = 324/15625


  • #6
    eoins23456
    Registered Users, Registered Users 2 Posts: 603 ✭✭✭eoins23456


    q9(a) 4/52 * 4/51 * 44/50


    q9(b)(i) (3/5)^4*(2/5)^2 is correct as a specific order is given i.e he has to lose the second and fourth game.If there wasnt a specific order there would be 15 possible combinations of him losing two games so you would multiply your answer by 6cr4 or 15.

    b(ii) wins exactly four games would 6cr4*(3/5)^4*(2/5)^2=4860/15625=972/3125

    b(iii)loses at least four games would be the probability he loses 4 games 6cr2*(3/5)^2*(2/5)^4+the probability he loses five -6cr1*(3/5)*(2/5)^5+the probability he loses them all (2/5)^6.If you add them all up 112/625

    c) z=(505-500)/(18/sqrt(36))=5/3=1.6 which is less then 1.96 so you dont reject the null hypothesis,conclusion is result consistent with companies claim

    at the back of the edco exam papers,well these ones are fairly old so it might have change it says1.66>1.645 so reject null hypothesis.this indicates you should use a one tailed test but i think you are supposed to use a two tailed test.


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  • #7
    krisityfer
    Closed Accounts Posts: 21 ✭krisityfer


    eoins23456 wrote: »
    If there wasnt a specific order there would be 15 possible combinations of him losing two games so you would multiply your answer by 6cr4 or 15.
    Ahhhh, that clears it up. Thanks for your help guys, here's hoping Paper 2 ain't a bitch.

    Edit* Although I still wouldn't say no if anyone happens to find the marking schemes, just in case I have problems with 2000-1997.


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