Question

A room has three lamp sockets. From a collection of 10 light bulbs of which only 6 are good, three bulbs are selected at random and placed in the sockets. What is the probability that will be light in the room?
$\left( A \right)\dfrac{{29}}{{30}}$
$\left( B \right)\dfrac{1}{6}$
$\left( C \right)\dfrac{5}{6}$
$\left( D \right)\dfrac{1}{{30}}$

Answer 1

Hint – In this particular question use the concept that to select (r) object out of (n) object we use combination rule (i.e. ${}^n{C_r}$), so first of all calculate the total number of outcomes to select 3 bulbs out of 10 bulbs, then separate good and faulty bulbs and make the cases according to the given condition so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given data:
There are three lamp sockets in a room.
Now there are 10 light bulbs in which only 6 are good and rest are faulty.
Therefore faulty bulbs = (10 – 6) = 4 bulbs
Now three bulbs are chosen at random out of 10 bulbs.
So the total number of ways to select three bulbs out of 10 bulbs = ${}^{10}{C_3}$ .
So the total number of outcomes = ${}^{10}{C_3}$.
Now we have to find the probability so that there is a light in the room.
So there are certain cases which satisfy the given condition.
$\left( a \right)$ When all the good bulbs are chosen.
So the number of ways to choose 3 good bulbs out of 6 good bulbs are ${}^6{C_3}$
$\left( b \right)$ When two good bulbs and one faulty bulb are chosen.
So the number of ways to choose 2 good bulbs out of 6 good bulbs and one faulty bulb out of 4 faulty bulbs are ${}^6{C_2} \times {}^4{C_1}$
$\left( c \right)$ When one good bulb and two faulty bulbs are chosen.
So the number of ways to choose 1 good bulb out of 6 good bulbs and two faulty bulbs out of 4 faulty bulbs are ${}^6{C_1} \times {}^4{C_2}$
So the total number of favorable number of outcomes = ${}^6{C_3}$ + ${}^6{C_2} \times {}^4{C_1}$ + ${}^6{C_1} \times {}^4{C_2}$
Now as we know that the probability (P) is the ratio of the favorable number of outcomes to the total number of outcomes so we have,
$\Rightarrow P = \dfrac{{{\text{favorable number of outcomes}}}}{{{\text{total number of outcomes}}}}$
Now substitute the values we have,
$\Rightarrow P = \dfrac{{{}^6{C_3}\; + \;{}^6{C_2} \times {}^4{C_1} + {}^6{C_1} \times {}^4{C_2}}}{{{}^{10}{C_3}}}$
Now as we know that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ so use this property we have,
$\Rightarrow P = \dfrac{{\dfrac{{6!}}{{3!\left( {6 - 3} \right)!}} + \dfrac{{6!}}{{2!\left( {6 - 2} \right)!}} \times \dfrac{{4!}}{{1!\left( {4 - 1} \right)!}} + \dfrac{{6!}}{{1!\left( {6 - 1} \right)!}} \times \dfrac{{4!}}{{2!\left( {4 - 2} \right)!}}}}{{\dfrac{{10!}}{{3!\left( {10 - 3} \right)!}}}}$
Now simplify this equation we have,
$\Rightarrow P = \dfrac{{\dfrac{{6!}}{{3!\left( 3 \right)!}} + \dfrac{{6!}}{{2!\left( 4 \right)!}} \times \dfrac{{4!}}{{1!\left( 3 \right)!}} + \dfrac{{6!}}{{1!\left( 5 \right)!}} \times \dfrac{{4!}}{{2!\left( 2 \right)!}}}}{{\dfrac{{10!}}{{3!\left( 7 \right)!}}}}$
$\Rightarrow P = \dfrac{{\dfrac{{6.5.4.3!}}{{3.2.1\left( 3 \right)!}} + \dfrac{{6.5.4!}}{{2.1\left( 4 \right)!}} \times \dfrac{{4.3!}}{{1\left( 3 \right)!}} + \dfrac{{6.5!}}{{1\left( 5 \right)!}} \times \dfrac{{4.3.2!}}{{2.1\left( 2 \right)!}}}}{{\dfrac{{10.9.8.7!}}{{3.2.1\left( 7 \right)!}}}}$
$\Rightarrow P = \dfrac{{20 + 15 \times 4 + 6 \times 6}}{{120}} = \dfrac{{116}}{{120}} = \dfrac{{29}}{{30}}$
So this is the required answer.
Hence option (A) is the correct answer.

Note – Whenever we face such types of question the key concept we have to remember is that always recall the formula of the combination rule (i.e. ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$), and also recall that the probability is the ratio of the favorable number of outcomes to the total number of outcomes, so first calculate these outcomes as above then put these outcomes in the formula of the probability and simplify we will get the required probability.