$\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} $

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

- 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}}$

Ask Question

Take Test

x

JEE MAIN, CBSE, AIPMT Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...