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Question #41816

Difficulty: 8/10 | 🟢 Free

The Question

A biased tetrahedral die has four faces coloured red, black, white, and white. The probability of rolling the red face is $\frac{1}{4}$ , the black face is $\frac{1}{4}$ , and a white face is $\frac{1}{2}$ . If the die is rolled $5$ times, let $p$ be the probability that either the red face appears at least twice, or the black face appears at least twice (or both). Find the value of $p$ .

Options

A.
$\frac{21}{32}$
B.
$\frac{47}{64}$
C.
$\frac{11}{32}$
D.
$\frac{5}{64}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Binomial Distribution Tail-Ends|Complementary Counting|Venn Diagram Overlap Logic

Why This Question Is Tricky

Common Mistakes

This question tests the application of complementary counting and the Inclusion-Exclusion Principle within a multinomial probability framework.

The primary trap lies in treating the two tail-end binomial events (at least 2 reds, at least 2 blacks) as mutually exclusive or independent.

Students often blindly calculate $P(E_1) + P(E_2)$ using the standard binomial formula $1 - P(X < 2)$ , yielding the trap answer $\frac{47}{64}$ , failing to realize that the events overlap and are negatively correlated (since they draw from the same finite number of trials).

The optimal path requires recognizing that the complement $P(E_1^c \cap E_2^c)$ forms a highly constrained, easily calculable sample space, avoiding the complex double-counting of the Venn diagram intersection.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Probability - Class 12
Concept	Binomial Distribution Tail-Ends
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free