Results 1 to 10 of 10

Math Help - Probability 1

  1. #1
    Member
    Joined
    Mar 2008
    Posts
    204

    Probability 1

    The word APE are written on the three cards, with one letter on each card. The three card are then shuffled and dealt face up in a line to form " word". Q1. List the possible equally likely "words" that could be so formed

    Q2. List the probability that the "word" so formed:
    (a) is the APE
    (b) is not the word PEA
    (c) starts with an A
    (d) does not start with E
    (e) has A and E as immediate neighbours in any order

    thank u for your time
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    The word APE can only be arranged in 6 ways, so that kind of boils it down.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Mar 2008
    Posts
    204
    Q1. List the possible equally likely "words" that could be so formed
    ans: 3! = 6 i.e APE , AEP, PAE, PEA, EAP, EPA

    Q2. List the probability that the "word" so formed:
    (a) is the APE
    P (APE) = 1/6

    (b) is not the word PEA
    P( not PEA) = 5/6

    (c) starts with an A
    P( start with A) = 2/6 = 1/3

    (d) does not start with E
    P( not start with E) = 4/6 = 2/3

    (e) has A and E as immediate neighbours in any order
    ???--stuck--what it mean by 'immediate neighbours in any order'

    so hope some one can reconfirm my answers on Q2 (a) to (d) and help me on guestion (e)

    Tq again
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jan 2009
    Posts
    34
    (a) to (d) all look good to me. As for (e) it is basically asking how many words have P as the middle letter, which I am sure you can figure out
    Last edited by mattty; January 21st 2009 at 08:30 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Mar 2008
    Posts
    204
    Quote Originally Posted by mattty View Post
    (a) to (d) all look good to me. As for (e) it is basically asking how many words have P as the middle letter, which I am sure you can figure out
    for your info the ans for Q2e is is 2/3. i also blur according to the books ans.so i think the ans is 1/3. Do u agree??

    and for
    2(e) has A and E as immediate neighbours in any order

    P ( A and E as immediate neighbours) = 2/6 = 1/3
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Jan 2009
    Posts
    34
    I am fairly sure that the answer for 2(e) is 1/3, yes. however, if your book says the answer is 2/3, then a second opinion wouldn't go astray in here
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    The solution to 2e is 2/3. A and E must be next to one another. That can be AE or EA. Count up the cases like this out of the 6. There are 4.

    4/6=2/3.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Jan 2009
    Posts
    34
    i think i'll stick to asking questions
    i read 2(e) as if it was asking for times when one of the letters had an a and e as immediate neighbours. (ape and epa 2/6 =1/3)
    Last edited by mattty; January 21st 2009 at 09:10 AM. Reason: clarification
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Mar 2008
    Posts
    204
    mattty & galactus...

    i am veery sorry .. it my mistake on what i'm writing here on 2e ans..

    for 2(c) the ans is 2/3 (our ans 1/3)
    2(e) the ans is 1/3 (our ans 1/3)

    i'm verry sorry on that...

    so we have the correct ans for 2(e) but not for 2(c).. but i think our ans on 2(c) is correct i.e 2/3...

    sorry for inconvenience
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Mar 2008
    Posts
    204
    any one agree the correct i.e 2c is 2/3...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 10
    Last Post: January 21st 2011, 11:47 AM
  2. Replies: 3
    Last Post: May 29th 2010, 07:29 AM
  3. fundamental in probability & convergence with probability 1
    Posted in the Advanced Statistics Forum
    Replies: 3
    Last Post: February 23rd 2010, 09:58 AM
  4. Replies: 1
    Last Post: February 18th 2010, 01:54 AM
  5. Replies: 3
    Last Post: December 15th 2009, 06:30 AM

Search Tags


/mathhelpforum @mathhelpforum