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Question #40557

Difficulty: 8/10 | 🟢 Free

The Question

Four distinct balls are to be distributed into three distinct boxes $B_1, B_2,$ and $B_3$ . For each ball, a fair coin is tossed. If it shows heads, the ball is placed in $B_1$ . If it shows tails, a fair die is rolled; if the die shows a number less than 5, the ball is placed in $B_2$ , otherwise it is placed in $B_3$ . What is the probability that no box remains empty?

Options

A.
$\frac{4}{9}$
B.
$\frac{1}{3}$
C.
$\frac{5}{18}$
D.
$\frac{1}{5}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Fundamental Principle of Counting (FPC)|Conditional Probability Constraints|Equiprobable Outcomes|Equiprobability Fallacy

Why This Question Is Tricky

Common Mistakes

This question targets the 'Equiprobability Fallacy' in combinatorial probability.

A student relying purely on the Fundamental Principle of Counting might calculate the total number of distributions as $3^4 = 81$ and the favorable distributions (none empty) as $3! \times S(4,3) = 36$ , incorrectly concluding the probability is $36/81 = 4/9$ .

This assumes all 81 outcomes are equally likely.

However, the conditional constraints (coin toss and die roll) create a non-uniform probability space for the boxes.

The correct approach requires recognizing that the individual permutations are not equiprobable and modeling the independent trials using the multinomial distribution or the inclusion-exclusion principle.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Permutations and Combinations - Class 11
Concept	Fundamental Principle of Counting (FPC)
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free