Question

A room has three electric lamps. From a collection of 10 electric bulbs of which 6 are good,3 are selected at random and put in the lamps. Find the probability that the room is lighted.
(a) $\dfrac{29}{30}$
(b) $\dfrac{1}{30}$
(c) $\dfrac{23}{30}$
(d) $\dfrac{7}{30}$

Answer 1

In order to solve this problem, we need to assume that the room is considered lighted even with one bulb working. Therefore, we need to find the probability of at least one bulb working. Also, we need to use the relation of combinations. The formula for the number of possible ways the $r$ thing is don’t out of $n$ things is given by ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ . The formula for the probability of is given by $\text{=}\dfrac{\text{desired outcome}}{\text{total possible outcome}}$ .

Complete step by step answer:
We need to find the probability that the room is lighted.
For that, we need to select 3 bulbs from the collection of 10 bulbs,
According to the condition, only 6 bulbs are in a working condition.
For this, we need to assume that if one of the lamps is lighted then the room will be lighted.
To get the room lighted we need at least one bulb to work from the three which we are selecting.
To solve this we will first find the probability that we pick all the three bulbs faulty.
So, there are 4 bulbs out of 10 which are no working.
The formula for the number of possible ways the $r$ thing is don’t out of $n$ things is given by ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$ .
Therefore the number of ways 3 bulbs can be picked out of 4 is given by ${}^{4}{{C}_{3}}$ .
Total number of possible ways that 3 bulbs are picked out of 10 is given by ${}^{10}{{C}_{3}}$ .
The formula for the probability of is given by $\text{=}\dfrac{\text{desired outcome}}{\text{total possible outcome}}$ .
Therefore the probability of the room is not lighted is $P=\dfrac{{}^{4}{{C}_{3}}}{{}^{10}{{C}_{3}}}$ .
Solving, we get,
$\begin{aligned} & P=\dfrac{\dfrac{4!}{3!\left( 4-3 \right)!}}{\dfrac{10!}{3!\left( 10-3 \right)!}}=\dfrac{4!\times 7!}{10!}=\dfrac{4\times 3\times 2\times 7\times 6\times 5\times 4\times 3\times 2}{10\times 9\times 8\times 7\times 6\times 5\times 4\times 3\times 2} \\\ & \Rightarrow P=\dfrac{1}{30}..........................(i) \\\ \end{aligned}$
The total probability is always 1.
Therefore, the probability that the room will be lighted is 1 – P
Solving we get,
$\Rightarrow 1-P=1-\dfrac{1}{30}=\dfrac{29}{30}$ .
Therefore, the probability that the room will be lighted is $\dfrac{29}{30}$ .
Hence, the correct option is (a).

Note: We cannot find directly the probability that the room is lighted. First, I need to find the probability that the room is not lighted and then get subtracted by 1. We need to be careful while calculating the factorial of the number. Also, we can cross-check that the probability of any event is always less than 1.