Thread: Binomial Distribution Probability

1. Binomial Distribution Probability

A toy manufacturing company tests the quality of the toys it manufactures. On a daily basis, 20 toys are taken at random for testing and to check that 95% meet the toy specification. The day's production will be accepted provided that no more than 2 toys fail to meet the specification standard.

n=20
Success(P)=0.75
Failure(Q)=0.25

=(^20 C 19)(0.75)^19(0.25)^20−19

=0.0211

The result seems lower than what I would expect. Where have I gone wrong?
Thanks.

2. Re: Binomial Distribution Probability

I will use the binomial coefficient notation. $_n C_r = \dbinom{n}{r}$.

Probability that there are zero failed toys:

$\dbinom{20}{0}(0.25)^0(0.75)^{20} = \dfrac{3486784401}{1099511627776} \approx 0.003$

Probability there is exactly one failed toy:
$\dbinom{20}{1}(0.25)^1(0.75)^{20-1} = \dfrac{5811307335}{274877906944} \approx 0.211$ (as you found)

Probability there are exactly two failed toys:
$\dbinom{20}{2}(0.25)^2(0.75)^{20-2} = \dfrac{36804946455}{549755813888} \approx 0.0669$

Probability that the day's manufactured toys pass inspection:
$\dfrac{3486784401}{1099511627776}+\dfrac{58113073 35}{274877906944}+\dfrac{36804946455}{549755813888 } \approx 0.09126$

3. Re: Binomial Distribution Probability Originally Posted by MikeS A toy manufacturing company tests the quality of the toys it manufactures. On a daily basis, 20 toys are taken at random for testing and to check that 95% meet the toy specification. The day's production will be accepted provided that no more than 2 toys fail to meet the specification standard.
n=20
Success(P)=0.75
Failure(Q)=0.25
To be quite honest. I have no idea what you are doing.
$1 - \sum\limits_{k = 0}^2 {\left[ {\dbinom{20}{k}} \right]{{\left( {.25} \right)}^k}{{\left( {.75} \right)}^{20 - k}}}$ is the probability
that no more than two will fail. SEE HERE

4. Re: Binomial Distribution Probability Originally Posted by Plato To be quite honest. I have no idea what you are doing.
$1 - \sum\limits_{k = 0}^2 {\left[ {\dbinom{20}{k}} \right]{{\left( {.25} \right)}^k}{{\left( {.75} \right)}^{20 - k}}}$ is the probability
that no more than two will fail. SEE HERE
I believe this gives the probability that more than two will fail.

5. Re: Binomial Distribution Probability Originally Posted by SlipEternal I believe this gives the probability that more than two will fail.
The sum gives the probability that none fails, one fails, or two fail. The complement of which is the probability that no more than two fail.

6. Re: Binomial Distribution Probability Originally Posted by Plato The sum gives the probability that none fails, one fails, or two fail. The complement of which is the probability that no more than two fail.
The complement of (none, one, or two fails) is MORE than two fail, not NO more than two. The sum of the probabilities that zero fails, one fails, or two fails IS the probability that no more than two fail.

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