Question

A carton contains 25 bulbs, 8 of which are defective. What is the probability that if a sample of 4 bulbs is chosen, exactly 2 of them will be defective?

Answer 1

because we don't know which 8 are defective so it can be 2 defective bulbs by 4

Answer 2

Probability

Let D be the event in which defective bulbs are chosen and N be  the event in which non-defective bulbs are chosen.

Total number of bulbs $=25$

Number of defective bulbs$=8$

We know that the probability of an event E

is given by P(E)= $\frac{Number \ of \ favourable \ outcomes}{Total \ number \ of \ outcomes}$

Given that a sample of 4 bulbs is chosen at random.

Total number of possible outcomes=$\binom{25}{4}$

We have to select 2 defective bulbs from 8 defective bulbs and 2 undefective bulb from 17 undefective bulbs.

So, the number of favourable outcomes is

$\binom{8}{2}\times \binom{17}{2}$

Therefore, the required probability =

$\frac{\binom{8}{2}\times \binom{17}{2}}{\binom{25}{4}}$

$=\frac{1904}{6325}$

Hence, it is the required probability.