Results 1 to 5 of 5

Math Help - conditional probability

  1. #1
    Newbie
    Joined
    Nov 2007
    Posts
    3

    conditional probability

    I'm having trouble solving a problem. I've been staring at it for almost an hour now, and have had hardly any progress...

    I'm trying to do problems 1 and 4 from here.

    Thanks in advance.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,546
    Thanks
    539
    Hello, valosn!

    Here's #1 . . .


    1. A factory produces items in boxes of two.
    Over the long run:
    . . (A): 92% of the boxes contain 0 defective items,
    . . (B): 5% of the boxes contain 1 defective item, and
    . . (C): 3% of the boxes contain 2 defective items.

    A box is picked at random from production, then an item is picked
    at random from that box. Given that the item is defective, what is
    the probability that the second item in the box is defective?

    We are concerned with two events:
    . . 1D: the first item is defective, and
    . . 2D: the second item is defective.

    We want: . {\color{blue}P(2D\,|\,1D) \;=\;\frac{P(1D \wedge 2D)}{P(1D)}} .[1]


    The numerator is: P(1D \wedge 2D), the probability that both are defective ... case (C).
    . . We are told that this happens: 3\% of the time.
    Hence: . {\color{blue}P(1D \wedge 2D) \:=\:\frac{3}{100}}


    The denominator is: P(1D), the probability that the first is defective.
    This can happen in two ways:

    (i) Case (B) which happens \frac{5}{100} of the time
    . . and we pick the defective item, probability \frac{1}{2}.
    Thus, this has probability: \frac{5}{100}\cdot\frac{1}{2} \:=\:\frac{1}{40}

    (ii) Case (C) which happens \frac{3}{100} of the time
    . . and we pick a defective item, probability 1.
    Thus, this has probability: \frac{3}{100}(1) \:=\:\frac{3}{100}

    Hence: . P(1D) \:=\:\frac{1}{40} + \frac{3}{100} \quad\Rightarrow\quad{\color{blue}P(1D) \:=\:\frac{11}{200}}


    Substitute into [1]: . P(2D\,|\,1D) \;=\;\frac{\dfrac{3}{100}}{\dfrac{11}{200}} \;=\;\boxed{\frac{6}{11}}

    Follow Math Help Forum on Facebook and Google+

  3. #3
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Here's 4

    recall that we say two events are independent if knowledge that one occurs does not change the probability of the other occurring.

    (a) here, there are different numbers of each type of ball in both boxes. however, the numbers are proportional. no matter which box we select, the probability of selecting a black ball from that box is the same, and the probability of selecting a red ball from that box is the same. thus, yes, the color of the ball is independent of which box is chosen. verify this using the formulas for conditional probability and show that you get the same result for selecting, say, a black ball given that you have selected box 1, and then, the probability of selecting a black ball given that you have selected box 2. do this for the red ball also

    (b) the events are now dependent. knowledge of which box we've selected changes the probability of a black ball being chosen. verify this using the formulas.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Nov 2007
    Posts
    3
    Thanks a lot! It was the denominator that was giving me trouble, but you explained it very clearly. Thank you!

    As for the 4th problem, I was having trouble with part b), but I figured it out.

    Again, thanks a lot for the help!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,383
    Thanks
    1474
    Awards
    1
    I have that book in front of me. I reviewed an early edition for Springer.
    For #4, first find P(R).
    \begin{array}{rcl}<br />
 P(R) & = & P(R \cap 1) + P(R \cap 2) \\ <br />
  & = & P(R|1)P(1) + P(R|2)P(2) \\ <br />
  & = & \frac{2}{5}\frac{1}{2} + \frac{8}{{20}}\frac{1}{2} \\ <br />
  & = & \frac{2}{5} \\   \end{array}

    Now we know that
    P(R|1) = \frac{2}{5}\frac{1}{2} = \frac{1}{5}\,\& \,P(R|2) = \frac{8}{{20}}\frac{1}{2} = \frac{1}{5}

    Do you think they are independent?
    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. Conditional Probability using the Law of Total Probability
    Posted in the Advanced Statistics Forum
    Replies: 1
    Last Post: October 7th 2010, 03:01 AM
  3. Continuous probability - conditional probability
    Posted in the Statistics Forum
    Replies: 1
    Last Post: October 1st 2009, 01:21 AM
  4. Conditional Probability
    Posted in the Statistics Forum
    Replies: 1
    Last Post: June 15th 2009, 04:14 AM
  5. Conditional probability.
    Posted in the Statistics Forum
    Replies: 4
    Last Post: March 4th 2009, 08:54 AM

Search Tags


/mathhelpforum @mathhelpforum