Question

A special lottery is to be held to select one student who will live in the only deluxe room in a hostel. There are 100 Year-III, 150 Year-II and 200 Year-I students who applied. Each Year-III student's name is placed in the lottery drum 3 times, each Year-II student's name 2 times, and each Year-I student's name 1 time. What is the probability that a Year-III student's name will be drawn?

• A. \(\frac{1}{8}\)
• B. \(\frac{2}{9}\)
• C. \(\frac{2}{7}\)
• D. \(\frac{3}{8}\)

Hint:

Total slips = (students x entries per student) added across all three groups; probability for Year-III = its slip count divided by the total.

Answer 1

The Correct Option is D

Step 1: Turn the entries into a simple ratio instead of raw totals.
The number of slips each group contributes is proportional to (number of students) times (entries per student). Write this as a ratio:
Year-III : Year-II : Year-I $= (100 \times 3) : (150 \times 2) : (200 \times 1) = 300 : 300 : 200$.

Step 2: Simplify the ratio.
Dividing every term by 100: $300:300:200 = 3:3:2$.

Step 3: Add the parts of the ratio.
Total parts $= 3 + 3 + 2 = 8$.

Step 4: Read off the Year-III share.
Year-III corresponds to 3 parts out of the total 8 parts, so its probability of being chosen is $\frac{3}{8}$.

Step 5: Cross-check against raw numbers.
The actual slip counts were 300, 300 and 200 out of 800 total, and $\frac{300}{800}$ also reduces to $\frac{3}{8}$, confirming the ratio shortcut gives the same result as working with the full totals.

Final Answer:
The probability that a Year-III student wins is $\frac{3}{8}$. \[ \boxed{\frac{3}{8}} \]