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It is known that every 100th TV produced is defective. If we choose sample of 10 TVs from the tracks, find:

a) probability that one TV is defective.

b) probability that at most two TVs are defective.

I don't know where to start from. Should I use Poisson distribution here or not?

p = 0.01

a) $f(1,10,p) = (10 1) p(1-p)^{10} = 10p(1-p)^{10}$

b) $f(2,10,p) = 10p(1-p)^{10} + (10 2) p^2(1-p)^8 = 10p(1-p)^{10} + 45 p^2(1-p)^8$

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Did you mean to say "It is known that, on average, one out of 100 produced TVs is defected."? –  Sasha Jul 3 '12 at 15:29
    
Is this homework? –  celtschk Jul 3 '12 at 15:31
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When you say "It is known that every 100th produced TV is defected", do you mean that the probability that a TV is defective is 1/100? If that is the case you treat the sample of 10 as a sample having a binomial distribution with p=0.01 and n=10. –  Michael Chernick Jul 3 '12 at 15:32
    
@Sasha yes. celtshsk no! –  Takarakaka Jul 3 '12 at 15:33
    
Many thanks! I have solved it now. @MichaelChernick –  Takarakaka Jul 3 '12 at 15:38
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2 Answers

up vote 4 down vote accepted

No you probably should not use the Poisson distribution as the Poisson is nothing more than the limiting case of the Binomial where n is large and p is small. So you usually shouldn't use Poisson when $np$ or $nq$ (that is, $n(1-p)$) is less than five.

As a clue:

  • If a mean or average probability of an event happening per unit time/per page/per mile cycled etc., is given, and you are asked to calculate a probability of n events happening in a given time/number of pages/number of miles cycled, then the Poisson Distribution is used.
  • If, on the other hand, an exact probability of an event happening is given, or implied, in the question, and you are asked to caclulate the probability of this event happening k times out of n, then the Binomial Distribution must be used.

in your question

Here we are given a definite probability, in this case, of defective components, $p = 0.01$ and hence $q = 0.99$ = Prob. not defective. Hence, Binomial, with $n = 10$. Expand $(q + p)^{10}$ to get

$q^{10} + 10 q^9 p + 10(10-1)/2! q^8 p^2 + \dots$

So $P(1) = 10 q^9 p = 0.09135$.

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Thank you for this information! @BYS2 –  Takarakaka Jul 3 '12 at 15:47
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The probability that one random TV is defected is $p=1/100$; the probability that it isn't is correspondingly $1-p$.

If we assume that the probability of one TV being defected is independent from the probability of another TV being defected (i.e. you only assume random failures, not e.g. a machine defect which makes a whole bunch of TVs defective at once), the peobabilities for different TVs just multiply. For example if you draw three TVs, the probability that the first TV is OK, the second defected and the third OK again is $(1-p)p(1-p)=9801/1000000$ or approximately 1%.

Now if you ask what is the probability that exactly $k$ TVs out of $n$ are defected, then you have several cases ecluding each other. For example, the first $k$ TVs could be defective, and all following are not. Or only the last $k$ are defective. Or the 1st, 3rd, 5th, …, $(2k-1)$th are defective. In total there are ${n \choose k} = \frac{n!}{k!(n-k)!}$ different possibilities. To get the total probability, you have to add up the probability for each single case. However in each single case, you have $k$ factors $p$ (because $k$ TVs are defective) and $n-k$ factors $1-p$ (because $1-p$ TVs are OK). Therefore your probability is just ${n\choose k}p^k(1-p)^{n-k}$. This is known as binomial distribution, because it's just one term in the binomial formula of $(p+q)^n$ where $q=1-p$ (which BTW immediately shows that the probabilities sum up to 1 as they should, because $(p+(1-p))^n=1^n=1$.

Now to your specific questions. You have a batch of 10 TVs, therefore $n=10$.

(a) You ask for the probability that one TV is defected. This can be interpreted in two ways: (a1) What is the probability that exactly one TV is defected? Or (a2) What is the probability that at least one TV is defected?

For (a1) it is now just a matter of inserting into the formula above. Since ${n \choose 1}=n$, we get $P_{a1}=10 p(1-p)^9 \approx 0.091 = 9.1\%$.

For case (a2), the probability that at least one TV is defected is one minus the probability that none is defected. But the probability for the latter is also easily calculated with the binomial distribution: Since ${n\choose 0}=1$, we get $P_{a2}=1-(1-p)^10\approx 0.096 = 9.6\%$

For case (b), note that at most two TVs are defected if either non, exactly one, or exactly two TVs are defected. Since those possibilities exclude each other (it's not possible that you have both exactly one and exactly two defected TVs in your sample), you just have to add those probabilities. Using ${n\choose 2}=\frac{n(n-1)}{2}$, we therefore get $P_b = (1-p)^10+10p(1-p)^9+45p^2(1-p)^8 \approx 0.9999 = 99.99\%$.

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Great article! @celtschk –  Takarakaka Jul 3 '12 at 16:20
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