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When bulbs are connected in series, the same current flows through each bulb. The total power consumed in a series circuit can be affected by the individual resistances of the bulbs.
Let's analyze the situation:
Step 1: Understand the power ratings of the bulbs.
- A installs a 200 W bulb.
- B installs a 50 W bulb.

Step 2: Convert the power ratings to resistances using the formula:
\( P = \frac{V^2}{R} \), thus, \( R = \frac{V^2}{P} \). Assuming the voltage (V) is the same for both bulbs, the resistances are:
- For A: \( R_A = \frac{V^2}{200} \)
- For B: \( R_B = \frac{V^2}{50} \)

Step 3: Calculate the total resistance when both are in series: \( R_{total} = R_A + R_B = \frac{V^2}{200} + \frac{V^2}{50} = \frac{V^2}{200} + \frac{4V^2}{200} = \frac{5V^2}{200} = \frac{V^2}{40} \).

Step 4: Find the current (I) through the circuit: \( I = \frac{V}{R_{total}} = \frac{V}{\frac{V^2}{40}} = \frac{40}{V} \).

Step 5: Analyze bulbs voltage drop:
Voltage across A's bulb: \( V_A = I R_A = \left(\frac{40}{V}\right) \left(\frac{V^2}{200}\right) = \frac{40V}{200} = \frac{V}{5} \).
Voltage across B's bulb: \( V_B = I R_B = \left(\frac{40}{V}\right) \left(\frac{V^2}{50}\right) = \frac{40V}{50} = \frac{4V}{5} \).

Step 6: Since A’s bulb has a higher resistance and a lower voltage drop across it, it will produce less light than intended despite being rated for more power, leading to inadequate lighting.

Therefore, A is more likely to fail the end-of-term examinations due to poor lighting conditions.