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Thread: Need help with binomial distribution using the nCr method.

  1. #1
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    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!
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  2. #2
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    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

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

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

    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 $\displaystyle \binom{5}{0}q^5p^0+\binom{5}{1}q^4p^1+\binom{5}{2} q^3p^2$

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

    It will be quite different to 0.61
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  3. #3
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    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

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

    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 $\displaystyle \binom{4}{1}p^3q^1=\binom{4}{3}p^3q^1$

    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 $\displaystyle \binom{4}{0}p^4q^0+\binom{4}{3}p^3q^1$

    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%
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