A lot contains 50 defective and 50 non-defective bulbs. Two bulbs are drawn at random, one at a time, with replacement. The events A, B and C are defined as follows:

A = (first bulb is detective)

B = (second bulb is non-defective)

C = (two bulbs are both defective or both non-defective)

Then

Let X = defective and Y = non-defective.

Then all possible outcomes are {XX, XY, YX, YY}

$\begin{array}{c}\mathrm{Also}, \mathrm{P}\left(\mathrm{XX}\right)=\frac{50}{100}\times \frac{50}{100}=\frac{1}{4}\\ \mathrm{P}\left(\mathrm{XY}\right)=\frac{50}{100}\times \frac{50}{100}=\frac{1}{4}\\ \mathrm{P}\left(\mathrm{YX}\right)=\frac{50}{100}\times \frac{50}{100}=\frac{1}{4}\\ \mathrm{P}\left(\mathrm{YY}\right)=\frac{50}{100}\times \frac{50}{100}=\frac{1}{4}\end{array}$

Here, $\mathrm{A}=\left(\mathrm{XX}\right)\cup \left(\mathrm{XY}\right); \mathrm{B}=\left(\mathrm{XY}\right)\cup \left(\mathrm{YY}\right); \mathrm{C}=\left(\mathrm{XX}\right)\cup \left(\mathrm{YY}\right)$

$\begin{array}{l}\therefore \mathrm{P}\left(\mathrm{A}\right)=\mathrm{P}\left(\mathrm{XX}\right)+\mathrm{P}\left(\mathrm{XY}\right)=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\\ \therefore \mathrm{P}\left(\mathrm{B}\right)=\mathrm{P}\left(\mathrm{XY}\right)+\mathrm{P}\left(\mathrm{YX}\right)=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\end{array}$

Now, $\mathrm{P}(\mathrm{A}\cap \mathrm{B})=\mathrm{P}\left(\mathrm{XY}\right)=\frac{1}{4}=\mathrm{P}\left(\mathrm{A}\right)\times \mathrm{P}\left(\mathrm{B}\right)$

Thus, A and B are independent events.

$\mathrm{P}(\mathrm{B}\cap \mathrm{C})=\mathrm{P}\left(\mathrm{YX}\right)=\frac{1}{4}=\mathrm{P}\left(\mathrm{B}\right)\times \mathrm{P}\left(\mathrm{C}\right)$

Thus, B and C are independent events.

$\mathrm{P}(\mathrm{C}\cap \mathrm{A})=\mathrm{P}\left(\mathrm{XY}\right)=\frac{1}{4}=\mathrm{P}\left(\mathrm{C}\right)\times \mathrm{P}\left(\mathrm{A}\right)$

Thus, C and A are independent events.

$\begin{array}{c}\mathrm{P}(\mathrm{A}\cap \mathrm{B}\cap \mathrm{C})=0\text{ (impossible events) }\\ \ne \mathrm{P}\left(\mathrm{A}\right)\times \mathrm{P}\left(\mathrm{B}\right)\times \mathrm{P}\left(\mathrm{C}\right)\end{array}$

Thus, A, B and C are dependent events.

Therefore, we can conclude that A, B, C are pairwise independent but A, B, C are dependent.