$\begin {array} ((A) 10^1 \quad & (B) \bigg(\frac{1}{2} \bigg)^5 \quad & (C) \bigg(\frac{9}{10} \bigg)^5 \quad & (D) \frac{9}{10} \end{array} $

- 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.

Given in a box containing n=100 bulbs, 10 are defective. We need to calculate the probability that out of a sample of $(n=5)$ bulbs, none is defective. Let the number of non-defective bulbs in this sample be X.

It is a Bernoulli 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 bulb) $=p=\large\frac{10}{100} = \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 $(q = 1 – p)$

P (none of bulbs are defective) $ = P (X=0) = \large^{5}C_0. \frac{1}{10}^0.\frac{9}{10}^{5–0} $$ = \large(\frac{9}{10})^5$

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