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The IEM Kolkata placement report for 2024 is not out yet. However the NIRF report for 2025 has been released.Check out the tabulated data given below to know more about IEM Kolkata placement 2024:

NOTE: The above data is obtained from the NIRF report 2025

The fee structure at NIILM University includes multiple components like tuition, application, and hostel fees. For the 2025 intake, the total tuition fees are INR 1.2 Lacs for the MBA Dual, INR 68,000 for the Executive MBA, and INR 35,000 for the PGDBM course. These figures are sourced from the official website and are subject to change.

28. The sample space of given condition is

$S=\{1,2,3,4,5,6\}$

Let,  $P(A)=$ probability of getting an odd number in first throw

$\Rightarrow P(A)=\frac{3}{6}=\frac{1}{2}$

$P(B)=$ probability of getting an even number

$\Rightarrow P(B)=\frac{3}{6}=\frac{1}{2}$

Probability of getting an even number in three times $=\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=\frac{1}{8}$

So, Probability of getting an odd number at least once

$=1-$ probability of getting an odd number in no throw

$=1-\frac{1}{8}$

$=\frac{8-1}{8}$

$=\frac{7}{8}$

$\therefore$ Probability of getting an odd number at least once $=\frac{7}{8}$

To be eligible for admission into the MBA at NIILM University, candidates must have completed a bachelor's degree from a recognised institution. A minimum of 45% marks is required for general category students. Candidates must ensure they meet the eligibility criteria before applying, as meeting academic requirements is crucial for consideration in the admission process.

A total of 3,032 students have been placed during the IEM Kolkata placements in the past three years. Check out the table below to see the breakdown of students placed in each year:

NOTE: The above data is obtained from the NIRF report 2025

27. Given,

$P(A)=0.3$

$P(B)=0.6$

$\mathrm{i.\; A}$ and $B$ are independent

$\Rightarrow P(A\cap B)=P(A).P(B)$

$=0.3\times 0.6$

$=0.18$

$\mathrm{ii.\; P}$ ( $A$ and not $B$ )

$\Rightarrow P(A\cap B^{\prime })=P(A)-P(A\cap B)$

$=0.3-0.18$

$=0.12$

$\mathrm{iii.\; P}$ ( $A$ or $B$ )

$\Rightarrow P(A\cup B)=P(A)+P(B)-P(A\cap B)$

$=0.3+0.6-0.18$

$=0.72$

$\mathrm{iv.\; P}$ (neither $A$ nor $B$ )

$\Rightarrow P(A^{\prime }\cap B^{\prime })=P(A\cup B{)}^{\prime }$

$=1-P(A\cup B)$

$=1-0.72$

$=0.28$

St. Clair College is among the top choices of international students for continuing higher education in Canada. St. Clair offers over 120 degree programs at UG and PG levels for its international students. St. Clair College tuition fee can vary from undergraduate to postgraduate programs. International students can refer to the program-specific page at St. Clair College for detailed information about the particular programs. Moreover, the top programs 1-year tuition fee at St. Clair College is listed below:

As of now, the Makhanlal Chaturvedi National University has not announced any specific application deadline for the BNTech programme. However, admission for the academic session 2025 is open. Candidates can refer to the official website and complete the application form to apply for their desired course. Additionally, it is crucial for applicants to check the academic qualifications before applying for the programme.

The highest package in IEM Kolkata has not been disclosed yet. Once released it will be updated here. Meanwhile, students can refer to the median package given in the table below:

26. Given,

$P(A)=\frac{1}{2}$

$P(B)=\frac{7}{12}$

$P$ (not $A$ or not $B$ )= $\frac{1}{4}$

$\Rightarrow P(A^{\prime }\cup B^{\prime })=P(A\cap B{)}^{\prime }$

$\Rightarrow P(A\cap B{)}^{\prime }=\frac{1}{4}$

$\Rightarrow 1-P(A\cap B)=\frac{1}{4}$

$\Rightarrow P(A\cap B)=1-\frac{1}{4}$

$\Rightarrow \frac{4-1}{4}$

$\Rightarrow \frac{3}{4}$

Now,

$P(A).P(B)=\frac{1}{2}\times \frac{7}{12}$

$=\frac{7}{24}\ne \frac{3}{4}$

$\Rightarrow P(A).P(B)\ne P(A\cap B)$

Therefore, $A$ and $B$ are not independent.

For the General category, a score of 370 or above is usually considered a good IPMAT Rohtak score. The total score in IPMAT Rohtak is 480.  While a score in the 300-350 range can still provide a strong chance of getting through, candidates should for at least 370 and above to increase their chances of being shortlisted for the further selection process.

Check out the table below to know more about IPMAT Rohtak scores:

Students can refer to the placements details given below as per the NIRF report 2025. IEM Kolkata has not released its placement report yet.

 NOTE: The above data is obtained from the NIRF report 2025

25. Given,

$P(A)=\frac{1}{4}$

$P(B)=\frac{1}{2}$

$P(A\cap B)=\frac{1}{8}$ Here,  $P$  (not $A$ not $B$ ) $=P(A^{\prime }\cap B^{\prime })=P(A\cup B{)}^{\prime }$

$\Rightarrow 1-P(A\cup B)$

$\Rightarrow 1-\left[P(A)+P(B)-P(A\cap B)\right]$

$\Rightarrow 1-\left[\frac{1}{4}+\frac{1}{2}-\frac{1}{8}\right]$

$\Rightarrow 1-\left[\frac{2+4-1}{8}\right]$

$\Rightarrow 1-\frac{5}{8}=\frac{8-5}{8}$

$\Rightarrow \frac{3}{8}$

IEM Kolkata has been providing successful placements over the years. NIRF Report for 2025 has been released. Check out the table below to know the key highlights of the IEM Kolkata placements 2024:

NOTE: The above data is obtained from the NIRF report 2025

After B.Com, apart from MBA, M.Com, CA, and CS, you can explore options like CFA, CMA, ACCA, Digital Marketing, Data Analytics, Banking & Finance diplomas, Law (LLB), UPSC or SSC exams, and entrepreneurship. These courses and career paths open opportunities in finance, taxation, corporate sectors, government jobs, and emerging fields like analytics and digital business.

IEM Kolkata provides good placements to its graduating students. The students are given excellent placement support every year placing them in top companies. Check out the table below to know the key highlights of the IEM Kolkata placements 2024:

NOTE: The above data is obtained from the NIRF report 2025

24. Given,

$P(A)=0.3$

$P(B)=0.4$

$A$ and $B$ are independent

$\mathrm{(i)\; P}(A\cap B)=P(A).P(B)$

$\begin{array}{l}=0.3\times 0.4\\ =0.12\end{array}$

$\mathrm{(ii)\; P}(A\cup B)=P(A)+P(B)-P(A\cap B)$

$\begin{array}{l}P(A\cup B)=0.3+0.4-0.12\\ =0.58\end{array}$

$\mathrm{(iii)\; P}(A/B)=\frac{P(A\cap B)}{P(B)}=\frac{0.12}{0.4}=0.3$

$\mathrm{(iv)\; P}(B/A)=\frac{P(B\cap A)}{P(A)}=\frac{0.12}{0.3}=0.4$