Question

A room has three lamp sockets. From a collection of 10 light bulbs of which only six are good, three bulbs are selected at random and placed in the sockets. What is the probability that will be light in the room ?

• A. $\frac{29}{30}.$
• B. $\frac{1}{6}.$
• C. $\frac{5}{6}.$
• D. $\frac{1}{30}.$

Answer 1

Answer: (A)

$\frac{29}{30}.$

Total number of bulbs $=10$
Number of ways of selecting $3$ bulbs out of $10$ bulbs is ${}^{10}{C}_3$
Now, given $6$ bulbs are good and $4$ are defective.
$\text{Probability of not having light in the room}=\text{Probability of selecting 3 defective bulbs}$
Probability of not having light in the room $=\frac{{}^4{C}_3}{{}^{10}{C}_3}=\frac{1}{30}$
So, probability of having light in the room is $1-\frac{1}{30}=\frac{29}{30}$