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Question #40369

Difficulty: 8/10 | 🟢 Free

The Question

A number $x$ is chosen at random from the set $S = \{1, 2, 3, \dots, 361\}$ . Let $E_1$ be the event that $x$ is a perfect square, and $E_2$ be the event that $x$ is a multiple of $19$ . If $E_1 \cup E_2$ occurs, a biased coin with probability of heads $\frac{20}{37}$ is tossed. If $E_1^c \cap E_2^c$ occurs, a different biased coin with probability of heads $\frac{5}{54}$ is tossed. Given that the coin toss results in heads, what is the probability that the chosen number $x$ was neither a perfect square nor a multiple of $19$ ?

Options

A.
$\frac{3}{5}$
B.
$\frac{2}{5}$
C.
$\frac{5}{8}$
D.
$\frac{3}{8}$

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

Cardinality Inclusion-Exclusion|De Morgan’s Laws in Probability|Bayes' Theorem Reversal

Why This Question Is Tricky

Common Mistakes

This question tests the synthesis of set-theoretic combinatorics (Inclusion-Exclusion) with conditional probability (Bayes' Theorem).

The primary trap lies in the denominator of the Bayes' reversal: students often fail to recognize that $E_1^c \cap E_2^c$ is the exact complement of $E_1 \cup E_2$ via De Morgan's Laws, or they miscalculate the intersection $|E_1 \cap E_2|$ by assuming independence instead of using number theory (a perfect square divisible by a prime $p$ must be divisible by $p^2$ ).

The use of $361$ ( $19^2$ ) forces a precise cardinality evaluation before the conditional probability can be correctly applied, punishing students who attempt to bypass the sample space restriction.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Sets - Class 11
Concept	Cardinality Inclusion-Exclusion
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free