Results 1 to 3 of 3

Math Help - Need help with binomial distribution using the nCr method.

  1. #1
    Newbie
    Joined
    Sep 2010
    Posts
    7

    Need help with binomial distribution using the nCr method.

    This is my homework question:

    There is an 81% chance a new Nogo car will have no defects. A dealer has just received 5 new Nogos. What is the probability that 2 or less of the cars have no defects?

    I worked the problem and got 61% is that right? If not, can you show me the right way to do it? I'm supposed to use the nCr formula.

    Here's the second question: again using the nCr method.

    18) A shipment of D-cell batteries left the Pretty Good Battery Company before the quality inspectors checked them. Unfortunately, 10 out of every 100 of these batteries was no good. Ferdinand bough 4 of these batteries from a local 24 hourstore.
    (a) What is the probability that exactly one of the batteries that he bought was
    no good? I computed this answer to be 42%. is this right?
    (b) What is the probability that at most one of the batteries that he bought was
    no good? I computed this and got 1.25%. I think that is very wrong...

    Thanks for all the help!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by turtle27 View Post
    This is my homework question:

    There is an 81% chance a new Nogo car will have no defects. A dealer has just received 5 new Nogos. What is the probability that 2 or less of the cars have no defects?

    I worked the problem and got 61% is that right? If not, can you show me the right way to do it? I'm supposed to use the nCr formula.

    Here's the second question: again using the nCr method.

    18) A shipment of D-cell batteries left the Pretty Good Battery Company before the quality inspectors checked them. Unfortunately, 10 out of every 100 of these batteries was no good. Ferdinand bough 4 of these batteries from a local 24 hourstore.
    (a) What is the probability that exactly one of the batteries that he bought was
    no good? I computed this answer to be 42%. is this right?
    (b) What is the probability that at most one of the batteries that he bought was
    no good? I computed this and got 1.25%. I think that is very wrong...

    Thanks for all the help!
    For the first question, you should be using the binomial expansion for

    (p+q)^5=\binom{5}{0}p^5q^0+\binom{5}{1}p^4q^1+\bin  om{5}{2}p^3q^2+\binom{5}{3}p^2q^3+\binom{5}{4}p^1q  ^4+\binom{5}{5}p^0q^5=1

    where p=0.81,\;\;\;q=0.19

    2 or less of the cars have no defects is...

    All 5 have defects, 4 cars have defects or 3 cars have defects....

    P(defective) = q

    Therefore the solution is \binom{5}{0}q^5p^0+\binom{5}{1}q^4p^1+\binom{5}{2}  q^3p^2

    Try that substituting in the values for "p" and "q".

    It will be quite different to 0.61
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by turtle27 View Post
    This is my homework question:

    Here's the second question: again using the nCr method.

    18) A shipment of D-cell batteries left the Pretty Good Battery Company before the quality inspectors checked them. Unfortunately, 10 out of every 100 of these batteries was no good. Ferdinand bough 4 of these batteries from a local 24 hourstore.
    (a) What is the probability that exactly one of the batteries that he bought was
    no good? I computed this answer to be 42%. is this right?
    (b) What is the probability that at most one of the batteries that he bought was
    no good? I computed this and got 1.25%. I think that is very wrong...

    Thanks for all the help!
    For your 2nd question...

    (a)

    You set it up as follows...

    P(good)=p=0.9

    P(defective)=q=0.1

    (p+q)=1\Rightarrow\ (p+q)^4=1=\binom{4}{0}p^4q^0+\binom{4}{1}p^3q^1+\b  inom{4}{2}p^2q^2+\binom{4}{3}p^1q3+\binom{4}{4}p^0  q^4

    This sums all of the exact probabilities in the situation.

    You are looking for the probability of exactly one defective out of the four.

    This is given by the term \binom{4}{1}p^3q^1=\binom{4}{3}p^3q^1

    for the given values of p and q.


    (b)

    At most one of the batteries being "no good" means "all 4 good" or "exactly 3 good" out of the 4 batteries bought.

    The probability of this is the sum of the binomial terms \binom{4}{0}p^4q^0+\binom{4}{3}p^3q^1

    again using the given values for p and q.

    For (a) you should get 0.2916 or 29.16%

    For (b) you should get .6561+0.2916=0.9477 or 94.77%
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: March 21st 2010, 05:25 PM
  2. Binomial Method
    Posted in the Business Math Forum
    Replies: 0
    Last Post: February 9th 2010, 12:28 PM
  3. Replies: 1
    Last Post: November 12th 2009, 12:38 AM
  4. Replies: 1
    Last Post: March 11th 2009, 11:09 PM
  5. Cumulative distribution function of binomial distribution
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: October 31st 2008, 03:34 PM

Search Tags


/mathhelpforum @mathhelpforum