FORUM  
 
 
 
 
     Recent Posts  |   Recent Topics  |   FAQ  |   Search  |   Forum Actions  |   Quick Links Prizes
 
Notices
 
Page 1 of 18 1 2 3 4 5 6 11 ... LastLast
Results 1 to 20 of 360
  1. #1
    Available on PM tusharsem's Avatar
    Join Date
    Sep 2008
    Posts
    1,923
    Thanks
    240
    Thanked 488 Times
    in 272 Posts

    Default Quant Thread for CAT 2009 (PnC & Probability)

    Here's a simple one :

    What is the probability of a year having 53 Sundays ?

    No options for this one.....

  2. #2
    Available on PM sumit2goody's Avatar
    Join Date
    Aug 2008
    Posts
    8,033
    Thanks
    856
    Thanked 917 Times
    in 558 Posts

    Default

    Quote Originally Posted by tusharsem View Post
    Here's a simple one :

    What is the probability of a year having 53 Sundays ?

    No options for this one.....
    Hi tushar i think this is a complex one..the simple one you are refeering to will be this:

    What is the probability of a leap year having 53 Sundays ?
    B-School Joiners,Please Click
    Moi Blog: (Taken a break) Dissociation
    Quant Thread RC ThreadVerbal Thread Leisure
    PGP Batch IIM Indore 2012

  3. #3
    Available on PM tusharsem's Avatar
    Join Date
    Sep 2008
    Posts
    1,923
    Thanks
    240
    Thanked 488 Times
    in 272 Posts

    Default

    Quote Originally Posted by sumit2goody View Post
    Hi tushar i think this is a complex one..the simple one you are refeering to will be this:

    What is the probability of a leap year having 53 Sundays ?

    Oh yes !! guess you are right...

    The ans to your ques would be 2/7 i think (can never be sure when its probability )

    And now just adding the options to the ques that I had posted earlier :

    What is the probability of an year having 53 Sundays ?

    A. 5/28
    B. 13/28
    C. 1/7
    D. 1/84
    E. None of these

  4. #4
    Available on PM sumit2goody's Avatar
    Join Date
    Aug 2008
    Posts
    8,033
    Thanks
    856
    Thanked 917 Times
    in 558 Posts

    Default

    Quote Originally Posted by tusharsem View Post
    Oh yes !! guess you are right...

    The ans to your ques would be 2/7 i think (can never be sure when its probability )

    And now just adding the options to the ques that I had posted earlier :

    What is the probability of an year having 53 Sundays ?

    A. 5/28
    B. 13/28
    C. 1/7
    D. 1/84
    E. None of these
    Well the answer is 5/28 but that is in the case when u dont consider the fact that 100, 200,300 are not leap years..in that case it will be very complex one..but if you neglect them and consider every fourth year as leap year than the answer is ..i will say 5/28
    B-School Joiners,Please Click
    Moi Blog: (Taken a break) Dissociation
    Quant Thread RC ThreadVerbal Thread Leisure
    PGP Batch IIM Indore 2012

  5. #5
    Invisible aaamresh's Avatar
    Join Date
    Aug 2008
    Posts
    1,236
    Thanks
    0
    Thanked 5 Times
    in 5 Posts

    Default Quant Thread for CAT 2009 (PnC & Probability)

    post permutation and combinations and probability questions here
    Last edited by sumit2goody; 10-Mar-09 at 12:57 PM. Reason: added PnC
    Solid sometimes Liquid but never Gaseous.

  6. #6
    Invisible aaamresh's Avatar
    Join Date
    Aug 2008
    Posts
    1,236
    Thanks
    0
    Thanked 5 Times
    in 5 Posts

    Default

    One on probability :

    Three cards are drawn from a well shuffled pack of 52 cards. Find the probability that all the three cards are of different suits and there is at least one face card.

    1) 1017/1105
    2) 1197/5505
    3) 129/1105
    4) 1197/5525
    5) 1017/5505

    originally posted by tusharsem........
    Solid sometimes Liquid but never Gaseous.

  7. #7
    Invisible aaamresh's Avatar
    Join Date
    Aug 2008
    Posts
    1,236
    Thanks
    0
    Thanked 5 Times
    in 5 Posts

    Default

    Quote Originally Posted by aaamresh View Post
    One on probability :

    Three cards are drawn from a well shuffled pack of 52 cards. Find the probability that all the three cards are of different suits and there is at least one face card.

    1) 1017/1105
    2) 1197/5505
    3) 129/1105
    4) 1197/5525
    5) 1017/5505

    originally posted by tusharsem........
    out of 52 cards 40 are number cards and 12 are face cards.

    case1:- out of 3, 2 are number cards and 1 is face card
    no of ways= 40C1*30C1*6C1=7200....( only 6 face cards are to be considerd to make all 3 cards of different colour)

    case2:- out of 3, 1 is number card and 2 are face cards
    no of ways= 40C1*9C1*6C1=2160

    case3:- all 3 are face cards
    no of ways= 12C1*9C1*6C1=648

    Hence total no of ways=7200+2160+648=10008

    no of ways of drawing 3 cards out of 52= 52C3=22100

    hence required prob.=10008/22100=1502/5525, which is not in the options...........
    where is the mistake i don't know. can anyone tell me my mistake.........???
    Solid sometimes Liquid but never Gaseous.

  8. #8
    Intern
    Join Date
    Feb 2009
    Locn
    Faridabad India
    Posts
    48
    Thanks
    0
    Thanked 1 Time in 1 Post

    Default

    Can sum1 post the method used for solving such problems? Or is the one stated here appropriate? If yes then where is the mistake?

  9. #9
    Available on PM sumit2goody's Avatar
    Join Date
    Aug 2008
    Posts
    8,033
    Thanks
    856
    Thanked 917 Times
    in 558 Posts

    Default

    Quote Originally Posted by aaamresh View Post
    One on probability :

    Three cards are drawn from a well shuffled pack of 52 cards. Find the probability that all the three cards are of different suits and there is at least one face card.

    1) 1017/1105
    2) 1197/5505
    3) 129/1105
    4) 1197/5525
    5) 1017/5505

    originally posted by tusharsem........
    I am doing it by other method

    total no. of cases are 52C3 for selecting 3 cards = 22100

    now p(at least 1 face card) = 1 - p(no face card)

    No. of case for no face card = 4C3 ( for selecting 3 suits) * 10 * 10 * 10 = 4000

    p(at least 1 face card) = 1- 4000/22100 = 18100/22100 = 181/221

    where am i wrong ????

    probability is liek thsi only u get different answers for different interpretations

    tusharsem post your method and answer
    B-School Joiners,Please Click
    Moi Blog: (Taken a break) Dissociation
    Quant Thread RC ThreadVerbal Thread Leisure
    PGP Batch IIM Indore 2012

  10. #10
    Available on PM tusharsem's Avatar
    Join Date
    Sep 2008
    Posts
    1,923
    Thanks
    240
    Thanked 488 Times
    in 272 Posts

    Default

    Quote Originally Posted by sumit2goody View Post
    I am doing it by other method

    total no. of cases are 52C3 for selecting 3 cards = 22100

    now p(at least 1 face card) = 1 - p(no face card)

    No. of case for no face card = 4C3 ( for selecting 3 suits) * 10 * 10 * 10 = 4000

    p(at least 1 face card) = 1- 4000/22100 = 18100/22100 = 181/221

    where am i wrong ????

    probability is liek thsi only u get different answers for different interpretations

    tusharsem post your method and answer


    Here is the solution :

    No of ways to select 3 cards from a pack of cards : 52C3

    Ways to select 3 suites out of 4 (this is because all the 3 cards have to be of different suites) : 4C3

    Now, P(to select at least 1 face card) =
    P(to select any card of that suit) - P(any non-face card of that suit)

    But since the 3 cards are drawn together, so we have

    (13C1)^3 - (10C1)^3

    Hence, the solution is :

    4C3*[(13C1)^3 - (10C1)^3] / (52C3)

    = 1197/5525 (phew !!!)

  11. #11
    Intern
    Join Date
    Feb 2009
    Locn
    Thane India
    Posts
    33
    Thanks
    0
    Thanked 1 Time in 1 Post

    Default

    Earlier I was getting sum weird answer. But then I solved by this method and got the rgt answer. Slowly I guess I will be back to solving such questions without much difficukty

  12. #12
    Master
    Join Date
    Feb 2009
    Locn
    Mumbai India
    Posts
    143
    Thanks
    0
    Thanked 2 Times
    in 2 Posts

    Default

    Yes that is right answer. Even I got same. Probability is difficult. Sumtimes 10 different people get 10 different answer. Sometimes even we will get 2 diffrent answer when we do sum 2 diffrent times

  13. #13
    Available on PM sumit2goody's Avatar
    Join Date
    Aug 2008
    Posts
    8,033
    Thanks
    856
    Thanked 917 Times
    in 558 Posts

    Default

    Quote Originally Posted by YadavAbhay View Post
    Yes that is right answer. Even I got same. Probability is difficult. Sumtimes 10 different people get 10 different answer. Sometimes even we will get 2 diffrent answer when we do sum 2 diffrent times
    thats the beauty of probability and thats why one needs to practice a lot in Prob
    B-School Joiners,Please Click
    Moi Blog: (Taken a break) Dissociation
    Quant Thread RC ThreadVerbal Thread Leisure
    PGP Batch IIM Indore 2012

  14. #14
    Invisible aaamresh's Avatar
    Join Date
    Aug 2008
    Posts
    1,236
    Thanks
    0
    Thanked 5 Times
    in 5 Posts

    Default

    Quote Originally Posted by tusharsem View Post
    Here is the solution :

    No of ways to select 3 cards from a pack of cards : 52C3

    Ways to select 3 suites out of 4 (this is because all the 3 cards have to be of different suites) : 4C3

    Now, P(to select at least 1 face card) =
    P(to select any card of that suit) - P(any non-face card of that suit)

    But since the 3 cards are drawn together, so we have

    (13C1)^3 - (10C1)^3

    Hence, the solution is :

    4C3*[(13C1)^3 - (10C1)^3] / (52C3)

    = 1197/5525 (phew !!!)
    ok tushar u got the right answer with a right approach.
    but still i m asking wat went wrong in my n sumit's method?
    itz also very necessary to clear our doubts regarding the way we approach for any question if we don't get the correct answer.
    Solid sometimes Liquid but never Gaseous.

  15. #15
    Available on PM sumit2goody's Avatar
    Join Date
    Aug 2008
    Posts
    8,033
    Thanks
    856
    Thanked 917 Times
    in 558 Posts

    Default

    Quote Originally Posted by aaamresh View Post
    ok tushar u got the right answer with a right approach.
    but still i m asking wat went wrong in my n sumit's method?
    itz also very necessary to clear our doubts regarding the way we approach for any question if we don't get the correct answer.
    i got the flaw in my method as i was doing 1 - 4C3 * 1000/22100 here actually i am subracting the case of different suits as well it should be as mentioned by tushar
    B-School Joiners,Please Click
    Moi Blog: (Taken a break) Dissociation
    Quant Thread RC ThreadVerbal Thread Leisure
    PGP Batch IIM Indore 2012

  16. #16
    Available on PM sumit2goody's Avatar
    Join Date
    Aug 2008
    Posts
    8,033
    Thanks
    856
    Thanked 917 Times
    in 558 Posts

    Default @Aamresh

    and in your method i think u need to choose the suit first like in ur case of choosing all 3 face cards

    without any restriction you can choose in 12C3 ways = 220

    but your method says no of ways= 12C1*9C1*6C1=648

    which is more than the total ways

    it should be 4C3 * 3C1 * 3C1 * 3C1 = 4*27 = 108
    Last edited by sumit2goody; 11-Mar-09 at 11:13 AM.
    B-School Joiners,Please Click
    Moi Blog: (Taken a break) Dissociation
    Quant Thread RC ThreadVerbal Thread Leisure
    PGP Batch IIM Indore 2012

  17. #17
    Invisible aaamresh's Avatar
    Join Date
    Aug 2008
    Posts
    1,236
    Thanks
    0
    Thanked 5 Times
    in 5 Posts

    Default

    Quote Originally Posted by sumit2goody View Post
    and in your method i think u need to choose the suit first like in ur case of choosing all 3 face cards

    without any restriction you can choose in 12C3 ways = 220

    but your method says no of ways= 12C1*9C1*6C1=648

    which is more than the total ways
    thanx a lot sumit................
    Solid sometimes Liquid but never Gaseous.

  18. #18
    Available on PM tusharsem's Avatar
    Join Date
    Sep 2008
    Posts
    1,923
    Thanks
    240
    Thanked 488 Times
    in 272 Posts

    Default

    Quote Originally Posted by aaamresh View Post
    ok tushar u got the right answer with a right approach.
    but still i m asking wat went wrong in my n sumit's method?
    itz also very necessary to clear our doubts regarding the way we approach for any question if we don't get the correct answer.

    Yes yes, I totally ralise that....actually am in office right now so thats why couldnt reply...its only now that I have got some time

    Think sumit has already recognised his mistake and pointed out yours too...

    But I will keep track of this the next time

  19. #19
    Available on PM tusharsem's Avatar
    Join Date
    Sep 2008
    Posts
    1,923
    Thanks
    240
    Thanked 488 Times
    in 272 Posts

    Default

    Guys !! Kindly help me with this question :

    When an n-digit number is selected from all n-digit nos (n <= 10), what is the probability that all its digits are distincts ?

    1) n! / 10^(n-1)
    2) 1 / 10^(n-1)
    3) (10 - n)! / 10^(n-1)
    4) 9! / (10 - n)! * (10^(n-1))

  20. #20
    Aficionado
    Join Date
    Feb 2009
    Locn
    Mumbai India
    Posts
    84
    Thanks
    0
    Thanked 1 Time in 1 Post

    Default

    How to solve this question? I am frightened of CAT now.

+ Reply to Thread
Page 1 of 18 1 2 3 4 5 6 11 ... LastLast

Similar Threads

  1. Quant Thread for CAT 2009 (Algebra)
    By baburaoapate in forum Quantitative Ability Forum
    Replies: 633
    Last Post: 25-Sep-14, 11:57 AM
  2. Quant Thread for CAT 2009 (Number System)
    By sumit2goody in forum Quantitative Ability Forum
    Replies: 2775
    Last Post: 15-Sep-11, 6:10 PM
  3. Quant Thread for CAT 2009 (Geometry n Mensuration)
    By StreetHawk in forum Quantitative Ability Forum
    Replies: 393
    Last Post: 11-Dec-09, 2:10 AM
  4. Quant Thread for CAT 2009 (Arithmetic)
    By sumit2goody in forum Quantitative Ability Forum
    Replies: 241
    Last Post: 06-Oct-09, 11:09 PM
  5. Quant Thread for CAT 2009 (Miscellaneous)
    By StreetHawk in forum Quantitative Ability Forum
    Replies: 133
    Last Post: 03-Oct-09, 9:02 PM

Bookmarks

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
 
 
 
 
 

 
 

 
©2008-2017   Enabilon Learning Private Limited. All rights reserved