Facebook Page
Twitter
RSS
+ Reply to Thread
Results 1 to 6 of 6
  1. MHB Apprentice
    veronica1999's Avatar
    Status
    Offline
    Join Date
    Jun 2012
    Location
    Jupimar
    Posts
    63
    Thanks
    78 times
    Thanked
    17 times
    #1
    An "unfair" coin has a 2/3 probability of turning up heads. If this coin is tossed 50 times, what is the probability that the total number of heads is even?

    I set up an equation but I am having trouble with the calculation.

    50C0 X (2/3)^0 X (1/3)^50 + 50C2 X (2/3)^2 X (1/3)^48 +............ 50C50 (2/3)^50X(1/3)^0

    I tried splitting it up thinking of the sigma notation.

    (2/3)^0 + (2/3)^2 .....................+ (2/3)^50 = { 1- (4/9)^25} / (1- 4/9)


    (1/3)^0 ......................... + (1/3)^50 = { 1- (1/9)^25}/ (1-1/9)

    It doesn't seem to work...

  2. Pessimist Singularitarian
    MHB Coder
    MHB Math Helper
    MHB Ambassador
    MarkFL's Avatar
    Status
    Offline
    Join Date
    Feb 2012
    Location
    St. Augustine, FL.
    Posts
    11,730
    Thanks
    32,038 times
    Thanked
    27,698 times
    Thank/Post
    2.361
    Trophies
    24 Highscores
    Awards
    MHB Statistics Award (2016)  

MHB Calculus Award (2016)  

MHB Pre-University Math Award (2016)  

MHB Model Helper Award (2015)  

MHB Calculus Award (2015)
    #2
    As you have realized, we need the binomial probability formula:

    $\displaystyle P(x)={n \choose x}p^x(1-p)^{n-x}$

    Identifying:

    $\displaystyle n=50$

    $\displaystyle p=\frac{2}{3}$

    we then need to compute:

    $\displaystyle P(X)=\sum_{k=0}^{25}P(2k)=\sum_{k=0}^{25}\left[{50 \choose 2k}\left(\frac{2}{3} \right)^{2k}\left(\frac{1}{3} \right)^{50-2k} \right]$

    Relying on technology, we find:

    $\displaystyle P(X)=\frac{358948993845926294385125}{717897987691852588770249}\approx0.5$

    Can you think of a way to obtain the approximation from the binomial theorem and the complementation rule?

  3. MHB Apprentice
    veronica1999's Avatar
    Status
    Offline
    Join Date
    Jun 2012
    Location
    Jupimar
    Posts
    63
    Thanks
    78 times
    Thanked
    17 times
    #3 Thread Author
    Quote Originally Posted by MarkFL View Post
    As you have realized, we need the binomial probability formula:

    $\displaystyle P(x)={n \choose x}p^x(1-p)^{n-x}$

    Identifying:

    $\displaystyle n=50$

    $\displaystyle p=\frac{2}{3}$

    we then need to compute:

    $\displaystyle P(X)=\sum_{k=0}^{25}P(2k)=\sum_{k=0}^{25}\left[{50 \choose 2k}\left(\frac{2}{3} \right)^{2k}\left(\frac{1}{3} \right)^{50-2k} \right]$

    Relying on technology, we find:

    $\displaystyle P(X)=\frac{358948993845926294385125}{717897987691852588770249}\approx0.5$

    Can you think of a way to obtain the approximation from the binomial theorem and the complementation rule?
    There is i/2 and 1/2( 1+ 1/3^50) in the answer choices. Which one would be the answer?
    i am trying to work backwards from the answer.

    oops, i see the second one is the answer
    Last edited by veronica1999; January 10th, 2013 at 15:48.

  4. Pessimist Singularitarian
    MHB Coder
    MHB Math Helper
    MHB Ambassador
    MarkFL's Avatar
    Status
    Offline
    Join Date
    Feb 2012
    Location
    St. Augustine, FL.
    Posts
    11,730
    Thanks
    32,038 times
    Thanked
    27,698 times
    Thank/Post
    2.361
    Trophies
    24 Highscores
    Awards
    MHB Statistics Award (2016)  

MHB Calculus Award (2016)  

MHB Pre-University Math Award (2016)  

MHB Model Helper Award (2015)  

MHB Calculus Award (2015)
    #4
    I see we aren't meant to either use technology, or use an approximation.

    Intuition tells us that since there are 26 favorable outcomes and 25 unfavorable that the probability will be slightly greater than 1/2. But how much greater?

    Perhaps if we rewrite the sum:

    $\displaystyle P(X)=\frac{1}{3^{50}}\sum_{k=0}^{25}\left[{50 \choose 2k}2^{2k} \right]$

    Now, let's let Y be the event of getting an odd number of heads, and write:

    $\displaystyle P(X)-P(Y)=\delta$

    $\displaystyle \frac{1}{3^{50}}\sum_{k=0}^{25}\left[{50 \choose 2k}2^{2k} \right]-\frac{1}{3^{50}}\sum_{k=0}^{24}\left[{50 \choose 2k+1}2^{2k+1} \right]=\delta$

    $\displaystyle 3^{50}\delta=\sum_{k=0}^{50}\left[{50 \choose k}2^{k}(-1)^{50-k}\right]$

    By the binomial theorem, we have:

    $\displaystyle 3^{50}\delta=(2-1)^{50}=1$

    and so:

    $\displaystyle \delta=\frac{1}{3^{50}}$

    Hence:

    $\displaystyle P(X)-P(Y)=\frac{1}{3^{50}}$

    $\displaystyle P(X)+P(Y)=1$

    Adding, we find:

    $\displaystyle 2P(X)=1+\frac{1}{3^{50}}$

    and so finally, we have:

    $\displaystyle P(X)=\frac{1}{2}\left(1+\frac{1}{3^{50}} \right)$

  5. MHB Apprentice
    veronica1999's Avatar
    Status
    Offline
    Join Date
    Jun 2012
    Location
    Jupimar
    Posts
    63
    Thanks
    78 times
    Thanked
    17 times
    #5 Thread Author
    Quote Originally Posted by MarkFL View Post
    I see we aren't meant to either use technology, or use an approximation.

    Intuition tells us that since there are 26 favorable outcomes and 25 unfavorable that the probability will be slightly greater than 1/2. But how much greater?

    Perhaps if we rewrite the sum:

    $\displaystyle P(X)=\frac{1}{3^{50}}\sum_{k=0}^{25}\left[{50 \choose 2k}2^{2k} \right]$

    Now, let's let Y be the event of getting an odd number of heads, and write:

    $\displaystyle P(X)-P(Y)=\delta$

    $\displaystyle \frac{1}{3^{50}}\sum_{k=0}^{25}\left[{50 \choose 2k}2^{2k} \right]-\frac{1}{3^{50}}\sum_{k=0}^{24}\left[{50 \choose 2k+1}2^{2k+1} \right]=\delta$

    $\displaystyle 3^{50}\delta=\sum_{k=0}^{50}\left[{50 \choose k}2^{k}(-1)^{50-k}\right]$

    By the binomial theorem, we have:

    $\displaystyle 3^{50}\delta=(2-1)^{50}=1$

    and so:

    $\displaystyle \delta=\frac{1}{3^{50}}$

    Hence:

    $\displaystyle P(X)-P(Y)=\frac{1}{3^{50}}$

    $\displaystyle P(X)+P(Y)=1$

    Adding, we find:

    $\displaystyle 2P(X)=1+\frac{1}{3^{50}}$

    and so finally, we have:

    $\displaystyle P(X)=\frac{1}{2}\left(1+\frac{1}{3^{50}} \right)$

    Thanks!!!
    Now it is absolutely clear.

  6. Pessimist Singularitarian
    MHB Coder
    MHB Math Helper
    MHB Ambassador
    MarkFL's Avatar
    Status
    Offline
    Join Date
    Feb 2012
    Location
    St. Augustine, FL.
    Posts
    11,730
    Thanks
    32,038 times
    Thanked
    27,698 times
    Thank/Post
    2.361
    Trophies
    24 Highscores
    Awards
    MHB Statistics Award (2016)  

MHB Calculus Award (2016)  

MHB Pre-University Math Award (2016)  

MHB Model Helper Award (2015)  

MHB Calculus Award (2015)
    #6
    I found this problem quite interesting and thought about it while I was out and so wanted to generalize a bit and cut out some of the unnecessary hand-waving...

    Let's say the probability of getting a heads is $\displaystyle p$ and we flip the coin $\displaystyle 2n$ times where $\displaystyle n\in\mathbb{N}$. What is the probability that the total number of heads is even?

    Let $\displaystyle P(X)$ be the probability that the total number of heads is even and $\displaystyle P(Y)$ be the probability that the total number of heads is odd.

    $\displaystyle P(X)+P(Y)=1$

    $\displaystyle P(X)-P(Y)=\sum_{k=0}^{2n}\left[{2n \choose k}p^{2n-k}(p-1)^k \right]=(2p-1)^{2n}$

    Adding, we find:

    $\displaystyle 2P(X)=1+(2p-1)^{2n}$

    $\displaystyle P(X)=\frac{1}{2}\left(1+(2p-1)^{2n} \right)$
    Last edited by MarkFL; January 10th, 2013 at 23:22.

Posting Permissions

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