Results 1 to 4 of 4

Math Help - conditional probablity

  1. #1
    Member
    Joined
    Aug 2007
    Posts
    148

    conditional probablity

    Hi everyone,

    Could someone please help me on this conditional probablity problem?

    Given:
    Coin tossing
    A balanced dime is tossed twice. The four possible equaly likely outcomes are HH,HT, TH, TT

    Let:

    A=event the first toss is heads
    B=event the second toss is heads
    C=event at least one toss is heads

    Determine the following probabilities and express your results in words. Compute the conditional probabilities directly; do not use the conditional probability rule

    a. P(B)

    Would it be 3/4???????

    b. P(B|A)

    Would it be 1/3????????

    c. P(B|C)

    Would this also be 1/3???????

    d. P(c)

    1/3????

    e. P(C|A)

    f. P(|not B))

    I really don't understand this. Could someone please explain this to me so that I can do the rest that are like this?

    any help would be greatly appreciated

    Thank you very much
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Site Founder MathGuru's Avatar
    Joined
    Mar 2005
    From
    San Diego
    Posts
    478
    Awards
    1
    For a)

    The possibilities are HH, HT, TH, TT

    so the probability that the second toss is heads is 2 out of the four possibilities. It makes sense because HT and TT are the only two that satisfy this condition. Therefore, 1/2
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Site Founder MathGuru's Avatar
    Joined
    Mar 2005
    From
    San Diego
    Posts
    478
    Awards
    1
    For d)

    3 out of the four possibilities have a H in them HT, TH, HH

    so the answer is 3/4.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Aug 2007
    Posts
    148
    Thank you very much

    I understand "d" now, but I still don't understand "a".

    For a)

    The possibilities are HH, HT, TH, TT

    so the probability that the second toss is heads is 2 out of the four possibilities. It makes sense because HT and TT are the only two that satisfy this condition. Therefore, 1/2
    How do you know that "HH" and "TT" are the only two that satisfy this condition? Wouldn't it be "HH" "HT" and "TH"? because they are the ones that have a "heads"

    Thank you
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: July 22nd 2011, 01:39 AM
  2. Replies: 2
    Last Post: March 3rd 2011, 10:05 PM
  3. Please help to find probablity
    Posted in the Algebra Forum
    Replies: 1
    Last Post: June 3rd 2010, 10:43 PM
  4. Conditional Probablity Problem
    Posted in the Statistics Forum
    Replies: 2
    Last Post: October 19th 2008, 02:33 PM
  5. Probablity
    Posted in the Math Topics Forum
    Replies: 4
    Last Post: June 15th 2008, 03:42 AM

Search Tags


/mathhelpforum @mathhelpforum