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
A) $\dfrac{3}{{64}}$
B) $\dfrac{1}{{16}}$
C) $\dfrac{9}{{64}}$
D) $\dfrac{5}{{38}}$

Answer 1

Firstly, find the total number of outcomes for selecting 3 bulbs out of 20 total bulbs.
Then, find the total number of outcomes favorable to the event that out of 3 bulbs in sample space, 2 are defective.
Thus, find the probability of the above event by using the formula $P(X) = \dfrac{{Number\,of\,outcomes\,favorable\,to\,event\,X}}{{Total\,number\,of\,outcomes}}$.

Complete step by step solution:
The sample space given here is 3 out of 20 bulbs i.e. we have to select 3 bulbs out of 20 bulbs.
So, the total number of outcomes is ${}^{20}{C_3} = \dfrac{{20!}}{{3! \times \left( {20 - 3} \right)!}} = \dfrac{{20!}}{{3! \times 17!}} = \dfrac{{20 \times 19 \times 18 \times 17!}}{{3 \times 2 \times 1 \times 17!}} = 1140$ .
Now, it is given that out of 20 bulbs, 5 are defective. So, the number of non-defective bulbs is 15.
Let X be the event that in the sample space of 3 bulbs, 2 bulbs are defective bulbs.
Now, the number of ways 2 bulbs out of 5 defective bulbs can be selected is ${}^5{C_2} = \dfrac{{5!}}{{2! \times 3!}} = \dfrac{{5 \times 4 \times 3!}}{{2 \times 1 \times 3!}} = 10.$ Also, the number of ways of selecting 1 bulb out of 15 non-defective bulbs is ${}^{15}{C_1} = \dfrac{{15!}}{{1! \times 14!}} = \dfrac{{15 \times 14!}}{{1 \times 14!}} = 15$ .
So, the number of favorable outcomes for event X is given by ${}^5{C_2} \times {}^{15}{C_1} = 10 \times 15 = 150$ outcomes.
Now, the probability of event X is given by $P(X) = \dfrac{{Number\,of\,outcomes\,favorable\,to\,event\,X}}{{Total\,number\,of\,outcomes}}$ .
$\therefore P(X) = \dfrac{{150}}{{1140}} = \dfrac{5}{{38}}$

So, option (B) is the correct answer.

Note:
Probability of any event is described by the chance of that event likely to happen. The probability of any event must be greater than or equal to 0 and lesser than or equal to 1 i.e. $0 \leqslant P(X) \leqslant 1$ .
The probability of any Universal event, for example the Sun rises from east, is always 1 and the probability of any impossible event, for example the Sun revolves around the earth, is always 0.