Question

From a lot of 10 bulbs, which includes 3 defectives, a sample of two bulbs is drawn at random . Find the probability distribution of the number of defective bulbs.

Answer 1

Hint: Two bulbs are drawn from a lot of 10 bulbs which has 7 non defective bulbs and 3 defective bulbs and the probability that both the bulbs drawn are non defective is given by $P(X = 0) = \dfrac{{{}^7{C_2}*{}^3{C_0}}}{{{}^{10}{C_2}}}$and the probability that one of the bulbs drawn is defective and the other is non defective is given by $P(X = 1) = \dfrac{{{}^7{C_1}*{}^3{C_1}}}{{{}^{10}{C_2}}}$ and the probability that both the bulbs drawn is defective is given by $P(X = 2) = \dfrac{{{}^7{C_0}*{}^3{C_2}}}{{{}^{10}{C_2}}}$. Tabulating these values we get the required probability distribution.

Complete step-by-step answer:
We are given that there are 3 defective bulbs and 7 non defective bulbs
Let X denote the random variable of defective bulbs
And X can take the values 0 , 1 and 2
Now for X = 0 , that is none of the bulbs drawn are defective we get the probability distribution to be
$ \Rightarrow P(X = 0) = \dfrac{{{}^7{C_2}*{}^3{C_0}}}{{{}^{10}{C_2}}}$
This tells us that the two bulbs drawn are from the set of 7 non defective bulbs and 0 bulbs are drawn from the set of 3 defective bulbs
$ \Rightarrow P(X = 0) = \dfrac{{\dfrac{{7*6}}{{1*2}}*1}}{{\dfrac{{10*9}}{{1*2}}}} = \dfrac{{42}}{{90}} = \dfrac{7}{{15}}$
Now for X = 1 , that is one of the bulbs drawn are defective we get the probability distribution to be
$ \Rightarrow P(X = 1) = \dfrac{{{}^7{C_1}*{}^3{C_1}}}{{{}^{10}{C_2}}}$
This tells us that the one bulb is drawn from the set of 7 non defective bulbs and 1 bulb is drawn from the set of 3 defective bulbs
$ \Rightarrow P(X = 1) = \dfrac{{7*3}}{{\dfrac{{10*9}}{{1*2}}}} = \dfrac{{21}}{{45}} = \dfrac{7}{{15}}$
Now for X = 2 , that both the bulbs drawn are defective we get the probability distribution to be
$ \Rightarrow P(X = 2) = \dfrac{{{}^7{C_0}*{}^3{C_2}}}{{{}^{10}{C_2}}}$
This tells us that the 0 bulbs drawn are from the set of 7 non defective bulbs and 2 bulbs are drawn from the set of 3 defective bulbs
$ \Rightarrow P(X = 2) = \dfrac{{1*\dfrac{{3*2}}{{1*2}}}}{{\dfrac{{10*9}}{{1*2}}}} = \dfrac{6}{{90}} = \dfrac{1}{{15}}$
Hence the distribution is given by

X	0	1	2
P(X)	$\dfrac{7}{{15}}$	$\dfrac{7}{{15}}$	$\dfrac{1}{{15}}$

Note: 1.Probability distributions indicate the likelihood of an event or outcome. Statisticians use the following notation to describe probabilities: p(x) = the likelihood that a random variable takes a specific value of x. The sum of all probabilities for all possible values must equal 1.
2.By using this condition when we add our probabilities we get $\dfrac{7}{{15}} + \dfrac{7}{{15}} + \dfrac{1}{{15}} = \dfrac{{15}}{{15}} = 1$