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It is known that 10% of certain articles manufactured are defective. What is the probability that in a random sample of 12 such articles, 9 are defective?

$\begin{array}{1 1}22\large\frac{9^3}{10^{11}} \\122\large\frac{9^3}{10^{3}} \\ 22\large\frac{9}{10}^{11} \\122\large\frac{9}{10}^{11} \end{array}$

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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.
10% of certain articles manufactured are defective. We need to determine the probability that in a random sample of $(n = 12)$ such articles, 9 are defective.
Let X be the number of times we select a defective article out of 12 articles. It is aBernoulli trial as they satisfy 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 (getting a defective article) $= p =$10%$\; = \large\frac{1}{10}$$\rightarrow q = 1 - p =\large \frac{9}{10} Since X has a bionomial distribution, the probability of x success in n-Bernoulli trials, P (X = x) = \large^{n}C_x. p^x.q^{n–x} where x = 0, 1, 2,...,n and (q = 1 – p) We need to calculate the probability of 9 items being defective, i.e, P (X = 9) \Rightarrow P (X=9) =\; ^{12}C_9 \times \large\frac{1}{10}^9$$\times \large\frac{9}{10}^{12-9}$$=\; \large\frac{12 \times\ 11 \times 10 \times 9!}{9! \times3 \times 2 \times 1}( \frac{1}{10})^9 (\frac{9}{10})^3$
$P (X = 9) = 22\large\frac{9^3}{10^{11}}$
answered Jun 21, 2013

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