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Question #41079

Difficulty: 8/10 | 🟢 Free

The Question

An urn contains 2 red balls and 2 black balls. Two balls are drawn at random from the urn without replacement. Let $X$ be the number of red balls drawn. A biased coin, whose probability of showing heads is $\frac{X+1}{4}$ , is then tossed 5 times. What is the probability that the coin shows heads at least 4 times?

Options

A.
$\frac{131}{512}$
B.
$\frac{179}{768}$
C.
$\frac{265}{1536}$
D.
$\frac{73}{768}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Multiplication Theorem|Sampling With/Without Replacement|Binomial Distribution Tail-Ends

Why This Question Is Tricky

Common Mistakes

This question tests the synthesis of Hypergeometric sampling (without replacement) and Binomial distribution tail-ends via the Multiplication Theorem.

A primary trap is assuming independent trials for the first stage (sampling with replacement), which leads to an incorrect binomial prior for $X$ and the trap answer 131/512.

Another common error is calculating the exact probability $P(Y=4)$ instead of the tail-end $P(Y \ge 4)$ , leading to 265/1536.

Students must correctly apply the Law of Total Probability across dependent stages and accurately sum the binomial tail.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Probability - Class 11
Concept	Multiplication Theorem
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free