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There are 5% defective items in a large bulk of items. What is the probability that a sample of 10 items will include not more than one defective item?

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• For any Binomial distribution $B (n, p),$ the probability of x success in n-Bernoulli trials, $P (X = x) = \large^{n}C_x. p^x.q^{n–x}$ where $x = 0, 1, 2,...,n$ and $(q = 1 – p)$
• Trials of a random experiment are called Bernoulli trials, if they satisfy the following conditions: (i) There should be a finite number of trials. (ii) The trials should be independent. (iii) Each trial has exactly two outcomes : success or failure. (iv) The probability of success remains the same in each trial.
The experiment is to remove an item from a sample of 10 and see if its defective or not. (n = 10). It is a case of Bernoulli trials as it satisfies the conditions (i) finite number of trials, (ii) independent trials, (iii) there is a definite outcome and (iv) the probability of success does not change for each trial.
P (drawing a defective item) = $p = 5$%$= \large\frac{5}{100} = \frac{1}{20}$
P (drawing a non-defective item) = $q = 1-p =1 - \large\frac{1}{20} = \frac{19}{20}$
X is a bionomial distribution. Here $n=10, p = \large\frac{1}{20},$ $q =\large \frac{19}{20}$.
Since X has a bionomial distribution, the probability of x success in n-Bernoulli trials, $P (X = x) = \large^{n}C_x. p^x.q^{n–x}$ where $x = 0, 1, 2,...,n$ and $(q = 1 – p)$
$\Rightarrow$ (P = not more than one defective item) = P (0 defective items) + P (1 defective items)
$\Rightarrow P (X = 0) = \large^{10}C_0$$\times \large\frac{1}{20}^0$$\times\large\frac{19}{20}^{10–0}$$= \large(\frac{19}{20})^{10} \Rightarrow P (X = 1) = \large^{10}C_1$$\times \large\frac{1}{20}^1$$\times\large\frac{19}{20}^{10–1}$$=10\times$$\large\frac{1}{20}$$\large(\frac{19}{20})^9$
$\Rightarrow P (X \leq1) = P (X = 0) + P (X = 1) = \large(\frac{19}{20})^{10} $$+ 10\times$$\large\frac{1}{20}$$\large(\frac{19}{20})^9 \Rightarrow P (X \leq 1) = \large(\frac{19}{20})^9$$\times(\large\frac{19}{20} + \frac{10}{20}) = \large\frac{29}{20} (\frac{19}{20})^9$

edited Oct 31, 2013 by pady_1