Question

A carton contains 20 bulbs, 5 of which are defective. The probability that, if a sample of 3 bulbs is chosen at random from the carton 2 will be defective is

Answer 1

Hint: In this question, first of all, identify the number of defective bulbs and the number of correct bulbs. Then find out the number of possible outcomes and the number of favorable outcomes of the event to get 2 defective balls and 1 undefective ball chosen randomly from 20 balls. So, use this concept to reach the solution to the given problem.

Complete step-by-step answer:
Given that is,
The number of bulbs a cartoon contains = 20
The number of defective bulbs in the cartoon = 5
The number of correct bulbs in the carton = number of bulbs in a carton – number of defective bulbs in the carton
\[ = 20{\text{ }}-{\text{ }}5 = 15\]
We know that the probability of an event \[E\] is given by \[P\left( E \right) = \dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}}\]
Given that a sample of 3 bulbs is chosen at random.
So, the total number of possible outcomes = \[^{20}{C_3}\]
We have to select 2 defective bulbs from 5 defective bulbs and 1 undefective bulb from 15 undefective bulbs.
So, the number of favourable outcomes is = \[{}^5{C_2} \times {}^{15}{C_1}\]
Therefore, the required probability = \[\dfrac{{{}^5{C_2} \times {}^{15}{C_1}}}{{{}^{20}{C_3}}} = \dfrac{5}{{38}}\].
Thus, the required probability is \[\dfrac{5}{{38}}\].

Note: The probability of an event is always lying between 0 and 1 i.e., \[0 \leqslant P\left( E \right) \leqslant 1\]. We know that the probability of an event \[E\] is given by \[P\left( E \right) = \dfrac{{{\text{Number of favourable outcomes}}}}{{{\text{Total number of outcomes}}}}\]. The total number of balls should be equal to the sum of defective balls and undefective balls.