Question

In an institute, an MBA exam has 4 sections and a sectional cutoff is applied. A candidate qualifies only if they clear \emph{every sectional cutoff. In how many ways may an applicant fail the exam?}

• A. 15
• B. 16
• C. 32
• D. 64

Hint:

Binary “pass/fail” setups are tailor-made for the complement rule: count all \(2^n\) patterns, then subtract the single “all-pass” configuration.

Answer 1

The Correct Option is A

Step 1: Encode outcomes.
For each of the four sections there are two outcomes: Pass (P) or Fail (F). Hence the sample space of section-wise results has \[ 2^4=16 \] possible patterns (think of 4-bit strings of P/F).
Step 2: Identify “qualify”.
To qualify, the candidate must pass all sections simultaneously. There is \emph{exactly one} pattern that satisfies this: \(\text{PPPP}\).
Step 3: Count “fail”.
Failing means “not all passes,” i.e., every pattern except PPPP. Thus \[ \#\text{(fail patterns)}=16-1=15. \] (Check via Inclusion–Exclusion).
Let \(A_i\) be the event “fails section \(i\)”. Then \[ |\cup A_i|=\sum |A_i|-\sum|A_i\cap A_j|+\sum|A_i\cap A_j\cap A_k|-|A_1\cap A_2\cap A_3\cap A_4|. \] Each \(A_i\) fixes one F and leaves the other three sections free \(\Rightarrow |A_i|=2^3=8\). Similarly, \(|A_i\cap A_j|=2^2=4\), \(|A_i\cap A_j\cap A_k|=2^1=2\), \(|A_1\cap A_2\cap A_3\cap A_4|=2^0=1\). Therefore, \[ \binom41\!8-\binom42\!4+\binom43\!2-\binom44\!1 =4\cdot8-6\cdot4+4\cdot2-1=32-24+8-1=15. \] This matches the complement count. \[ \boxed{15} \]