Results 1 to 7 of 7

Math Help - Need help on probability with biasedness! *TOUGH QTN*

  1. #1
    Newbie
    Joined
    Apr 2008
    Posts
    16

    Need help on probability with biasedness! *TOUGH QTN*

    Suppose now I have two BIASED coins in two containers (4 coins in total).
    Container 1's coins have a probability of L1 of getting heads and Container 2's coins have a probability of L2 of getting heads. L1 is not equal to L2.

    Which will give a higher chance of getting two heads if I:
    a) Select one container at random then toss the two coins in it
    b) Toss one coin each from each container

    Please guide me and give me hints. I want to come up with the solution myself!

    For a start, this is about conditional probability, right?
    Also, I think option A will give a higher chance, but I'm not sure if I'm correct.

    IF I were to use imaginary figures taking L1 as 0.9 and L2 as 0.4, by doing:
    a) Pr1 = (0.9 * 0.9) * 0.5 = 0.405 | Pr2 = (0.4 * 0.4) * 0.5 = 0.08
    b) Pr = 0.9 * 0.4 = 0.36 (< Pr1)
    Option A is better in this case

    If L1 is 0.5 and L2 is 0.3,
    a) Pr1 = (0.5 * 0.5) * 0.5 = 0.125 | Pr2 = (0.3 * 0.3) * 0.5 = 0.045
    b) Pr = 0.5 * 0.3 = 0.15 (> Pr1)
    Option B is better in this case

    After here, I'm stuck and unsure how to proceed next to justify my choice. Please help! ^^ Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by zepher View Post
    Suppose now I have two BIASED coins in two containers (4 coins in total).
    Container 1's coins have a probability of L1 of getting heads and Container 2's coins have a probability of L2 of getting heads. L1 is not equal to L2.

    Which will give a higher chance of getting two heads if I:
    a) Select one container at random then toss the two coins in it
    b) Toss one coin each from each container

    Please guide me and give me hints. I want to come up with the solution myself!

    For a start, this is about conditional probability, right?
    Also, I think option A will give a higher chance, but I'm not sure if I'm correct.

    [snip]
    a) Select one container at random then toss the two coins in it:

    Pr(2 heads) = Pr( 2 heads | container 1) Pr(container 1) + Pr( 2 heads | container 2) Pr(container 2) =  (L_1)^2 \left( \frac{1}{2} \right) + (L_2)^2 \left( \frac{1}{2} \right) = \frac{1}{2} ( (L_1)^2 + (L_2)^2 ).

    b) Toss one coin each from each container:

    Pr(2 heads) = Pr(head from container 1) Pr(head from container 2) = L_1 L_2.


    Consider the function Q(L_1, L_2) = \frac{1}{2} ( (L_1)^2 + (L_2)^2 ) - L_1 L_2 = \frac{1}{2} (L_1 - L_2)^2.

    You're interested in whether Q > 0 or Q < 0 .....
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Apr 2008
    Posts
    16
    Quote Originally Posted by mr fantastic View Post

    Consider the function Q(L_1, L_2) = \frac{1}{2} ( (L_1)^2 + (L_2)^2 ) - L_1 L_2 = \frac{1}{2} (L_1 - L_2)^2.

    You're interested in whether Q > 0 or Q < 0 .....
    Actually, what's the  \frac{1}{2} (L_1 - L_2)^2 , I don't really get why I must equate to that. Shouldn't I equate \frac{1}{2} ( (L_1)^2 + (L_2)^2 ) > L_1 L_2 to prove that A is a better choice than B? But even so, how should I go about proving it?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by zepher View Post
    Actually, what's the  \frac{1}{2} (L_1 - L_2)^2 , I don't really get why I must equate to that. Shouldn't I equate \frac{1}{2} ( (L_1)^2 + (L_2)^2 ) > L_1 L_2 to prove that A is a better choice than B? But even so, how should I go about proving it?
    If you looked more carefully at my reply you would note that:

    1. The function Q is the difference between the two probabilities given by option a) and option b).

    2.  \frac{1}{2} (L_1 - L_2)^2 is the simplification of this function and therefore of the difference. In other words, the difference between the two probabilities given by option a) and option b) is  \frac{1}{2} (L_1 - L_2)^2 .

    3. The difference is always greater than zero. Therefore the probability given by option a) is always greater than the probability given by option b) .......
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Apr 2008
    Posts
    16
    Oh I got it now! But how did you simplify to get \frac{1}{2} (L_1 - L_2)^2 ? I can't seem to simplify to that. Sorry, I'm a bit dumb here =P
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Apr 2008
    Posts
    16
    anybody can help? kinda urgent now =P
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by zepher View Post
    Oh I got it now! But how did you simplify to get \frac{1}{2} (L_1 - L_2)^2 ? I can't seem to simplify to that. Sorry, I'm a bit dumb here =P
    Q(L_1, L_2) = \frac{1}{2} ( (L_1)^2 + (L_2)^2 ) - L_1 L_2 = {\color{red}\frac{1}{2} ((L_1)^2 - 2L_1 L_2 + (L_2)^2)} = \frac{1}{2} (L_1 - L_2)^2.

    It's expected that you can factorise standard forms.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. tough probability distribution problem
    Posted in the Statistics Forum
    Replies: 3
    Last Post: November 29th 2009, 11:00 PM
  2. Super Tough Probability Question
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: November 26th 2009, 07:00 AM
  3. Probability of faulty parts shipped. Tough question!
    Posted in the Advanced Statistics Forum
    Replies: 4
    Last Post: May 16th 2009, 11:16 PM
  4. Biasedness & Consistency
    Posted in the Advanced Statistics Forum
    Replies: 5
    Last Post: April 28th 2009, 06:20 PM
  5. Need Help On Tough Conditional Probability Problem
    Posted in the Advanced Statistics Forum
    Replies: 11
    Last Post: April 21st 2008, 09:24 AM

Search Tags


/mathhelpforum @mathhelpforum