A lot of 100 bulbs from a manufacturing process is known to contain 10 defective and 90 non-defective bulbs. If a sample of 8 bulbs is selected at random, the probability that the sample has atleast one defective bulb is:

Concept:  

The binomial distribution is the discrete probability distribution that gives only two possible results in an experiment, either Success or Failure.

Bernoulli trials: An experiment in which N  trials are made of an event, with probability p of success in any given trial and probability q = 1 - p of failure in any given trial.

The binomial distribution gives the probability to obtain n number of successful trials which is shown below:

P = \(_{n}^{N}\textrm{C}\)(p)ⁿ(q)^N-n  = \(_{n}^{N}\textrm{C}\)(p)ⁿ(q)^N-n

Calculation:

Given:

There are 10 defective bulbs in a group of 100 bulbs.

⇒ Probability of drawing a defective bulb from the group of 100 bulbs =  \(\frac{10}{100}\) = 0.1

⇒ Probability of drawing a non-defective bulb from the group of 100 bulbs = \(\frac{90}{100}\) = 0.9

⇒ Probability of drawing at least one defective bulb needed to be found out when 8 bulbs are selected at random.

This is an example of Binomial distribution.

Here, N = 8, p = 0.1, q = 0.9.

Hence, the required probability using the binomial theorem is given by,

Probability of drawing at least 1 defective bulb = 1 - Probability of drawing no defective bulbs

\(\Rightarrow\) Probability of drawing at least 1 defective bulb = 1 - \(_{0}^{8}\textrm{C}\)(p)⁰(q)^8-0

\(\Rightarrow\) Probability of drawing at least 1 defective bulb = 1 - (0.1)⁰(0.9)⁸

\(\Rightarrow\) Probability of drawing at least 1 defective bulb = 1 - (\(\frac{9}{10}\))⁸

Hence, the correct answer is option 4.