Results 1 to 5 of 5

Thread: Probability-2

  1. #1
    Trying to make mom and pop proud
    Join Date
    Jun 2010
    Posts
    11
    Rep Power
    3


    Good post? Yes | No

    Probability-2

    A bag contains 5 blue and 4 black balls. 3 are drawn at random. What is the probability that 2 are blue and 1 is black?

    Ans with explantion pls.

    Thanks

  2. #2
    Eager!
    Join Date
    May 2010
    Posts
    70
    Rep Power
    3


    Good post? Yes | No
    Probability of picking Blue in first instance = 5/9
    Probability of picking Blue in second instance = 4/8=1/2
    Probability of picking Black in three instance = 4/7
    Probability of picking (B,B,Black) in 3 random picks = (5/9)*(1/2)*(4/7)
    = 10/63
    This sequence can be arranged in 3 ways:
    (B,B,Black), (B,Black,B), (Black,B,B)

    So, Probability = (10/63)*3 = 10/21

    Thanks and regards,
    Vivek

  3. #3
    Trying to make mom and pop proud
    Join Date
    Aug 2010
    Posts
    17
    Rep Power
    3


    Good post? Yes | No
    Thank you very much.

  4. #4
    Fight like a soldier
    Join Date
    Mar 2009
    Location
    New Delhi, India, India
    Posts
    182
    Rep Power
    5


    Good post? Yes | No
    Quote Originally Posted by viveksurya View Post
    Probability of picking Blue in first instance = 5/9
    Probability of picking Blue in second instance = 4/8=1/2
    Probability of picking Black in three instance = 4/7
    Probability of picking (B,B,Black) in 3 random picks = (5/9)*(1/2)*(4/7)
    = 10/63
    This sequence can be arranged in 3 ways:
    (B,B,Black), (B,Black,B), (Black,B,B)

    So, Probability = (10/63)*3 = 10/21

    Thanks and regards,
    Vivek
    Yeah this may be the answer .

    You have considered that the probability without ( Replacement) . But I feel that with replacement the answer may be different.

    Although the question doesn't clearly mention with replacement/without replacement , can we assume that the case to be without replacement when no data pertaining to with replacement/without replacement is given.

    Someone please help.

  5. #5
    Trying to make mom and pop proud
    Join Date
    Aug 2010
    Posts
    23
    Rep Power
    3


    Good post? Yes | No
    Quote Originally Posted by viveksurya View Post
    Probability of picking Blue in first instance = 5/9
    Probability of picking Blue in second instance = 4/8=1/2
    Probability of picking Black in three instance = 4/7
    Probability of picking (B,B,Black) in 3 random picks = (5/9)*(1/2)*(4/7)
    = 10/63
    This sequence can be arranged in 3 ways:
    (B,B,Black), (B,Black,B), (Black,B,B)

    So, Probability = (10/63)*3 = 10/21

    Thanks and regards,
    Vivek
    thank you .in the question it doesn't clearly mention that it is with replacement/without replacement , can we assume that this case to be without replacement when no data pertaining to with replacement/without replacement is given.

    Someone please help.

Thread Information

Users Browsing this Thread

There are currently 1 users browsing this thread. (0 members and 1 guests)

Similar Threads

  1. probability
    By astral glow in forum GRE Math
    Replies: 16
    Last Post: 06-21-2012, 02:09 PM
  2. probability
    By mbawannabe in forum GMAT Problem Solving
    Replies: 2
    Last Post: 04-30-2008, 08:33 PM
  3. Probability
    By kluevehe in forum GMAT Problem Solving
    Replies: 1
    Last Post: 04-28-2008, 06:14 AM
  4. Probability
    By grems in forum GMAT Problem Solving
    Replies: 2
    Last Post: 04-26-2008, 05:51 PM
  5. Probability
    By dahiya in forum GMAT Math
    Replies: 2
    Last Post: 06-15-2005, 06:17 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
  •  

SEO by vBSEO ©2010, Crawlability, Inc.