Results 1 to 4 of 4

Math Help - Zany probablity problem.....

  1. #1
    Senior Member
    Joined
    Jan 2009
    Posts
    296

    Zany probablity problem.....

    A lady had a biopsy on a possibly cancerous tumor. She instructs the doctor to flip a coin. If it comes up heads, will call if the news is good but not if it is bad. If it comes up tails, the doctor will not call no matter what the results are. Let a be the probability that the tumor is cancerous and b the conditional probability that the tumor is cancerous given the doctor does not call.

    Find b in terms of a and determine which probability is larger.


    Intuition tells me the probability of b is the larger of the two, but I don't know how to show this.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Jan 2009
    Posts
    296
    Nevermind I think I got it:

    I believe B to be bigger and after some calculations I got

    B= 2a/(1+a)....can anyone just confirm or reject this answer?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,811
    Thanks
    701
    Hello, zhupolongjoe!

    It took a couple of readings to get the informaton digested.


    A lady had a biopsy on a possibly cancerous tumor.
    She instructs the doctor to flip a coin.
    If it comes up Heads, he will call if the news is good, but not if it is bad.
    If it comes up Tails, the doctor will not call no matter what the results are.

    Let a be the probability that the tumor is cancerous
    and b the conditional probability that the tumor is cancerous, given the doctor does not call.

    Find b in terms of a and determine which probability is larger.

    We have: . \begin{array}{c}P(\text{cancer}) \:=\:a \\ P(\text{cancer }|\text{ not call}) \:=\:b  \end{array}


    Consider the probability that he does not call.
    There are two cases:

    (1) The coin turns up Heads and the tumor is cancerous.
    . . .This probability is: . \tfrac{1}{2}a

    (2) The coin turns up Tails, so he does not call.
    . . .This probability is: . \tfrac{1}{2}

    Hence: . P(\text{not call}) \:=\:\tfrac{1}{2}a + \tfrac{1}{2}


    From Bayes' Theorem, we have: . P(\text{cancer }|\text{ not call}) \;=\;\frac{P(\text{cancer }\wedge\text{ not call})}{P(\text{not call})}

    This becomes: . b \;=\;\frac{\frac{1}{2}a}{\frac{1}{2}a + \frac{1}{2}} \quad\Rightarrow\quad b \:=\:\frac{a}{a+1}

    \text{We have: }\;b \;=\;a\underbrace{\left(\frac{1}{a+1}\right)}_{\te  xt{less than 1}}\quad\hdots \text{ That is, }b\text{ equals }a\text{ times a number less than 1.}

    Therefore: . b \:<\:a

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Jan 2009
    Posts
    296
    Hmmm...that's almost what I have....

    I have

    P(heads/cancer)=1/2 a
    P(tails/cancer)=1/2 a
    P(tails/no cancer)=1/2(1-a)

    So

    B= .5a+.5a/(.5a+.5a+.5(1-a))
    =2a/(a+1)

    What is wrong with my reasoning?

    Thank you.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Beginning Probablity Problem
    Posted in the Statistics Forum
    Replies: 2
    Last Post: February 2nd 2011, 08:17 PM
  2. Probablity Problem
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: June 20th 2010, 05:25 PM
  3. Fractions in a probablity problem?
    Posted in the Algebra Forum
    Replies: 2
    Last Post: April 15th 2009, 02:22 PM
  4. Conditional Probablity Problem
    Posted in the Statistics Forum
    Replies: 2
    Last Post: October 19th 2008, 02:33 PM
  5. really hard probablity problem
    Posted in the Statistics Forum
    Replies: 2
    Last Post: October 30th 2007, 05:56 PM

Search Tags


/mathhelpforum @mathhelpforum