Results 1 to 14 of 14

Math Help - Randomly chosen family

  1. #1
    Junior Member
    Joined
    Sep 2011
    Posts
    26

    Randomly chosen family

    A number of children in the families is 0 - 4. Children and probability are:
    0 child = 0.15,
    1 = 0.25,
    2 = 0.3,
    3 = 0.2,
    4 = 0.1
    Question: What is the probability of this randomly chosen family having exactly 2 girls?

    I think we must consider total probability.
    - And there are 8 ways that out of four children 1 are girls 4!/(4!*1!)
    - 8 ways that out of four children 2 are girls 4!/(3!*2!)
    - 8 ways that out of four children 3 are girls 4!(2!*3!)
    - 5 ways that out of four children 4 are girls 4!(1!*4!)
    Boys and girls do not correlate, they are independent. Probability that there is a girl is 1/2. We choose randomly one family ( which could have 0 or 1 or 2 or 3 or 4 children ).


    So two girls in a randomly chosen family has over 30? possibilities
    And of these 0-4 children having families are exactly 2 girls
    probability is 4/?
    Unless this is 5/? - because 0,1,2,3,4 = 5 different alternatives.



    Some thoughts;
    I think this is not needed: 0.15 * 0.25 * 0.3 * 0.2 * 0.1 ...
    BUT: this could be what is needed here: P(B) = P(A) P(B|A) + P(A^c) P(B|A^c)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4

    Re: Randomly chosen family

    Quote Originally Posted by stevetall View Post
    A number of children in the families is 0 - 4. Children and probability are:
    0 child = 0.15,
    1 = 0.25,
    2 = 0.3,
    3 = 0.2,
    4 = 0.1
    Question: What is the probability of this randomly chosen family having exactly 2 girls?

    I think we must consider total probability.
    - And there are 8 ways that out of four children 1 are girls 4!/(4!*1!)
    - 8 ways that out of four children 2 are girls 4!/(3!*2!)
    - 8 ways that out of four children 3 are girls 4!(2!*3!)
    - 5 ways that out of four children 4 are girls 4!(1!*4!)
    Boys and girls do not correlate, they are independent. Probability that there is a girl is 1/2. We choose randomly one family ( which could have 0 or 1 or 2 or 3 or 4 children ).


    So two girls in a randomly chosen family has over 30? possibilities
    And of these 0-4 children having families are exactly 2 girls
    probability is 4/?
    Unless this is 5/? - because 0,1,2,3,4 = 5 different alternatives.



    Some thoughts;
    I think this is not needed: 0.15 * 0.25 * 0.3 * 0.2 * 0.1 ...
    BUT: this could be what is needed here: P(B) = P(A) P(B|A) + P(A^c) P(B|A^c)
    P(exactly 2 girls)=P(exactly two girls | 2 children)P(2 children)+P(exactly two girls | 3 children)P(3 children)+P(exactly two girls | 4 children)P(4 children)

    The conditional probabilities can be evaluated using the appropriate binomoal distributions.

    CB
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Sep 2011
    Posts
    26

    Re: Randomly chosen family

    Indeed, the conditional probabilities are the right tools here!

    Here is my new effort.

    P(A) = exactly 2 girls P(A^c)= 2 children P(B) = 3 children P(C) = 4 children

    P(A) = P(A | A^c ) P(A^c ) + P(A | B) P(B ) + P(A | C) P(C)

    And probabilitie were; ( so 0 child = 0.15 and 1 = 0.25 are not needed in this )

    2 children = 0.3,
    3 children = 0.2,
    4 children = 0.1

    So could it be:
    P(A) = P(A | 0.3 ) P(0.3 ) + P(A | 0.2) P(0.2 ) + P(A | 0.1) P(0.1)

    But I don't know how a solution should be done.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4

    Re: Randomly chosen family

    Quote Originally Posted by stevetall View Post
    Indeed, the conditional probabilities are the right tools here!

    Here is my new effort.

    P(A) = exactly 2 girls P(A^c)= 2 children P(B) = 3 children P(C) = 4 children

    P(A) = P(A | A^c ) P(A^c ) + P(A | B) P(B ) + P(A | C) P(C)

    And probabilitie were; ( so 0 child = 0.15 and 1 = 0.25 are not needed in this )

    2 children = 0.3,
    3 children = 0.2,
    4 children = 0.1

    So could it be:
    P(A) = P(A | 0.3 ) P(0.3 ) + P(A | 0.2) P(0.2 ) + P(A | 0.1) P(0.1)

    But I don't know how a solution should be done.
    Well you should be able to compute the first of these; the probability that if you have two children both are girls.

    But all of them are subject to the same method:

    P(N girls in M children)=b(N;M,0.5)

    where b(.;.,.) is the binomial distribution mass function.

    CB
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Sep 2011
    Posts
    26

    Re: Randomly chosen family

    So it was; P(A) = exactly 2 girls P(A^c)= 2 children P(B) = 3 children P(C) = 4 children

    P(A) = P(A | A^c ) P(A^c ) + P(A | B) P(B ) + P(A | C) P(C)

    and here's new.

    probability that in two children both are girls = 1/2 * 1/2 = 1/4

    probability that in three children two are girls = 1/2 * 1/2 * 1/2 = 1/6

    probability that in four children two are girls = 1/2 * 1/2 * 1/2 * 1/2 = 1/8
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4

    Re: Randomly chosen family

    Quote Originally Posted by stevetall View Post
    So it was; P(A) = exactly 2 girls P(A^c)= 2 children P(B) = 3 children P(C) = 4 children

    P(A) = P(A | A^c ) P(A^c ) + P(A | B) P(B ) + P(A | C) P(C)

    and here's new.

    probability that in two children both are girls = 1/2 * 1/2 = 1/4

    probability that in three children two are girls = 1/2 * 1/2 * 1/2 = 1/6

    probability that in four children two are girls = 1/2 * 1/2 * 1/2 * 1/2 = 1/8
    >>binomial distribution<<

    CB
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Sep 2011
    Posts
    26

    Re: Randomly chosen family

    I read it and try once again.

    So it was; P(A) = exactly 2 girls P(A^c)= 2 children P(B) = 3 children P(C) = 4 children

    P(A) = P(A | A^c ) P(A^c ) + P(A | B) P(B ) + P(A | C) P(C)

    and here's new.

    probability that in two children both are girls = 1/4 * 1/4 = 1/8

    probability that in three children two are girls = 1/4 * 1/4 * 1/4 =

    probability that in four children two are girls = 1/4 * 1/4 * 1/4 * 1/4 =
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Sep 2011
    Posts
    26

    Re: Randomly chosen family

    I found errors and I had forgot to put rest of the results. I am terribly sorry, but I have really tried to solve this.

    We had:

    P(A) = exactly 2 girls P(A^c)= 2 children P(B) = 3 children P(C) = 4 children

    P(A) = P(A | A^c ) P(A^c ) + P(A | B) P(B ) + P(A | C) P(C)

    And my solution is:

    probability that in two children both are girls = 1/4 + 1/4 = 1/8

    probability that in three children two are girls = 1/4 + 1/4 + 1/4 = 1/12

    probability that in four children two are girls = 1/4 + 1/4 + 1/4 + 1/4 = 1/16


    AND: probability that there are exactly 2 girls =

    P(A) = exactly 2 girls P(A^c)= 2 children P(B) = 3 children P(C) = 4 children

    P(A) = P(A | A^c ) P(A^c ) + P(A | B) P(B ) + P(A | C) P(C)

    I am not sure how to use this: P(N girls in M children)=b(N;M,0.5)
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4

    Re: Randomly chosen family

    Quote Originally Posted by stevetall View Post
    I found errors and I had forgot to put rest of the results. I am terribly sorry, but I have really tried to solve this.

    We had:

    P(A) = exactly 2 girls P(A^c)= 2 children P(B) = 3 children P(C) = 4 children

    P(A) = P(A | A^c ) P(A^c ) + P(A | B) P(B ) + P(A | C) P(C)

    And my solution is:

    probability that in two children both are girls = 1/4 + 1/4 = 1/8

    probability that in three children two are girls = 1/4 + 1/4 + 1/4 = 1/12

    probability that in four children two are girls = 1/4 + 1/4 + 1/4 + 1/4 = 1/16



    AND: probability that there are exactly 2 girls =

    P(A) = exactly 2 girls P(A^c)= 2 children P(B) = 3 children P(C) = 4 children

    P(A) = P(A | A^c ) P(A^c ) + P(A | B) P(B ) + P(A | C) P(C)

    I am not sure how to use this: P(N girls in M children)=b(N;M,0.5)
    The part in red is nonsense the part in blue is how to compute these.

    CB
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Junior Member
    Joined
    Sep 2011
    Posts
    26

    Re: Randomly chosen family

    b(N;M,0.5) - i will try.




    ( N over 0.5 ) =

    2 children:

    2!
    ____

    0.5!(2-0.5)!

    = 2/1 = 2





    3 children:

    3!
    ____

    0.5!(3-0.5)!

    = 6/2 = 3




    4 children:


    4!
    ____

    0.5!(4-0.5)!


    = 24/3 = 8
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Junior Member
    Joined
    Sep 2011
    Posts
    26

    Re: Randomly chosen family

    I give up. I don't know what a solution is and how that must be done, too difficult.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Junior Member
    Joined
    Sep 2011
    Posts
    26

    Re: Randomly chosen family

    But I want to tell you what I achieved.

    Let's mark a girl with g and a boy with b. Because there could be four children, then there have to be
    2 * 2 * 2 * 2 = 16

    gggg
    gggb
    ggbb
    ggbg

    bbbb
    bbbg
    bbgg
    bbgb

    bggg
    bggb
    bgbg
    bgbb

    gbgg
    gbgb
    gbbg
    gbbb

    I find 7 hits, so probability that there are two girls is 7/16. ??
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4

    Re: Randomly chosen family

    Quote Originally Posted by stevetall View Post
    b(N;M,0.5) - i will try.




    ( N over 0.5 ) =

    2 children:

    2!
    ____

    0.5!(2-0.5)!

    = 2/1 = 2





    3 children:

    3!
    ____

    0.5!(3-0.5)!

    = 6/2 = 3




    4 children:


    4!
    ____

    0.5!(4-0.5)!


    = 24/3 = 8
     b(N;M,p)={M \choose N} p^n(1-p)^{M-N}=\frac{M!}{N!(M-N)!}p^n(1-p)^{M-N}

    Alternative notation:

     b(N;M,p)={_MC_N}\; p^n(1-p)^{M-N}}=\frac{M!}{N!(M-N)!}p^n(1-p)^{M-N}

    b(2;2,0.5)=\frac{2!}{2!0!}(0.5)^2=0.25

    b(2;3,0.5)=\frac{3!}{2!(3-2)!}(0.5)^2(0.5)^1=3 \times 0.5^3=0.375

    b(2;4,0.5)=\frac{4!}{2!(4-2)!}(0.5)^2(0.5)^2=2 \times 3 \times 0.5^4=0.375

    See the first shaded block at the link given, also scroll down for a binomial calculator.

    CB
    Last edited by CaptainBlack; October 12th 2011 at 05:34 AM.
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Junior Member
    Joined
    Sep 2011
    Posts
    26

    Re: Randomly chosen family

    That was fantastic!, really great work. Thank you! Really interesting solution, it sure was too hard for me. But I really tried.

    And I visited that link again interesting - now I tried it myself.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 7
    Last Post: July 25th 2010, 07:26 PM
  2. Replies: 2
    Last Post: June 1st 2010, 02:49 AM
  3. Replies: 3
    Last Post: January 26th 2010, 10:08 AM
  4. Probability of being chosen
    Posted in the Statistics Forum
    Replies: 3
    Last Post: November 24th 2009, 04:38 PM
  5. probability that a randomly chosen member
    Posted in the Statistics Forum
    Replies: 1
    Last Post: November 2nd 2009, 02:09 PM

Search Tags


/mathhelpforum @mathhelpforum