Question

Rohit buys 12 bulbs out of which 6 are defective, his brother chooses 3 bulbs at random for three sockets in a room. Find the probability that the room is lighted.

Answer 1

Number of bulbs Rahul bought- 12
Number of defected bulbs-6
Number of bulbs his brother chooses-3
Number of bulbs left with him- 12-6-3=3
P(bulbs lighted)= 3/12= 1/4
Hence,the probability will be 1/4 in another words, its 25% and in decimal form it is 0.25

Answer 2

Answer: 0.125

Step-by-step explanation:

Given : Rohit buys 12 bulbs out of which 6 are defective.

Then, the proportion of white bulbs = $p=\dfrac{6}{12}=0.5$

Binomial probability formula to find the probability of getting success in x trials :

$P(x)=^nC_x p^x(1-p)^{n-x}$, where n is the sample size, p is the probability of success in each trial.

Now, if  bulbs at random for three sockets in a room, then the probability that none of them is defective :

$P(0)=^3C_0 (0.5)^0(1-0.5)^{3}\\\\=(1)(0.5)^3=0.125$

∴ The probability that the room is lighted=0.125