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Santa Clara University's MS in Computer Science is taught under the Department of Computer Science and Engineering. The Computer Science programme is inclined towards designing and implementing software and hardware, and solving real-world problems using effective methods. Students willing to take part in the course must have a background in Computer Science, Computer Engineering, or Computer Science and Engineering. 

Candidates can submit the hard/printed copy of Admission Form and relevant documents by hand or by post to

Office of the Director,

University Institute of Legal Studies,  

Panjab University, Sector–14, Chandigarh-160014 

40. Let, $E_1:$ event of choosing $Q$ headed coin

$E_2:$ event of choosing biased coin

$E_3:$ event of choosing unbiased coin

$P(E_1)=P(E_2)=P(E_3)=\frac{1}{3}$

Let $A$ be the event that shows head,

$P(A/E_1)=P$ (coin shows heads, given that it is a headed coin) $=1$

$P(A/E_2)=P$ (coin showing up head, given that it is biased coin) $=\frac{75}{100}=\frac{3}{4}$

$P(A/E_3)=P$ (coin showing head, given that it is unbiased coin) $=\frac{1}{2}$

The probability that the coin is two headed, given that it shows heads, is given by $P(E_1/A)$

Using Baye’s theorem,

$P(E_1/A)=\frac{P(E_1).P(A/E_1)}{P(E_1).P(A/E_1)+P(E_2).P(A/E_2)+P(E_3).P(A/E_3)}$

$=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\times 1}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\times 1+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\times \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\times \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}$

$=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\frac{4+3+2}{12}}=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$12$}\right.}=\frac{1}{3}\times \frac{12}{9}=\frac{4}{9}$

NIILM MBA selection process involves shortlisting candidates based on their academic performance in the bachelor's degree examination. For the Executive MBA, candidates must also have a minimum of three years of relevant work experience. After the initial shortlisting, the university may conduct document verification and counselling sessions to finalise admissions.

The university follows a holistic approach where it gives priority to certain academic and non-academic factors. If the students start working on the factors given below, their probability of getting into Santa Clara may increase:

Applicants must submit a set of documents which generally include their bachelor's degree certificate or provisional certificate, mark sheets, proof of age (such as birth certificate or class 10 mark sheet), passport-sized photographs, category certificate (if applicable), and any other documents specified by the university. It is important to check the official website for the complete and updated list before applying.

39. Let, $E_1:$ event person has disease

$E_2:$ event person has no disease

$A:$ be event that blood test is positive

As $E_1$ and $E_2$ are events which are complementary to each other,

Then, $P(E_1)+P(E_2)=1$

$P(E_2)=1-P(E_1)$

$P(E_1)=0.1\mathrm{\%}=\frac{0.1}{100}=0.001$

$\therefore$ $P(E_2)=1-0.001$

$=0.999$

$P(A/E_1)=P$ (result is positive given that person has disease) $=99\mathrm{\%}=0.99$

$P(A/E_2)=P$ (result is positive given that person has no disease) $=0.5\mathrm{\%}=0.005$

Now, the probability that person has a disease, given that his test result is positive is $P(E_1/A)$

By using Baye’s theorem,

$P(E_1/A)=\frac{P(E_1).P(A/E_1)}{P(E_1).P(A/E_1)+P(E_2).P(A/E_2)}$

$=\frac{0.001\times 0.99}{0.001\times 0.99+0.999\times 0.005}$

$=\frac{0.00099}{0.00099+0.004995}$

$=\frac{0.00099}{0.005985}=\frac{990}{5985}=\frac{110}{665}$

$\Rightarrow P(E_1/A)=\frac{22}{133}$

According to the official class profile of 2027, the expected GPAs from international students vary by department. In addition, the GPA expectation for the School of Engineering is between 3.7 and 3.97, which is quite high. Students can refer to the table below for the GPA requirements to get into Santa Clara:

Start by registering online on the official NIILM University admissions portal—create your account, verify your email, complete the application form with your academic and personal details, upload required documents, and pay the application fee. Once shortlisted, admissions proceed based on your merit, followed by counselling, document verification, and fee payment to confirm your seat.

There are multiple factors upon which the BTech seat intake of MCNUJC Bhopal depends. Therefore, the university offers a total of  40 seats for the BTech programme. Candidates can refer to the following key points that affect the BTech seat intake of MCNUJC Bhopal:

• Availability of seats
• Number of students who applied
• Number of students shortlisted
• Performance in the entrance exam

Santa Clara University uses the Common App for undergraduate admission and the official university portal for postgraduate admission. Students must fulfill the basic eligibility by holding a bachelor's degree and proof of high school completion for PG and UG studies, and submit all the necessary paperwork. The most crucial documents include academic transcripts, a letter of recommendation, a statement of purpose, a resume, standardized test scores, and English language requirements.

38. Let, $E_1$ :event that knows the answer

$E_2$ :event that he guesses the answer

$P(E_1)=\frac{3}{4}$ and $P(E_2)=\frac{1}{4}$

Let $A$ be the event that the answer is correct

Also, $P(A/E_1)=P$ (correct answer given that he knows) $=1$

$P(A/E_2)=P$ (correct answer given that he guess) $=\frac{1}{4}$

Now, probability that he knows the answer given that he answer it correctly is $P(E_1/A)$ ,

By Baye’s theorem,

$P(E_1/A)=\frac{P(E_1).P(A/E_1)}{P(E_1).P(A/E_1)+P(E_2).P(A/E_2)}$

$=\frac{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\times 1}{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\times 1+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\times \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}=\frac{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$16$}\right.}$

$=\frac{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\frac{12+1}{16}}=\frac{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\raisebox{1ex}{$13$}\!\left/ \!\raisebox{-1ex}{$16$}\right.}$

$=\frac{3}{4}\times \frac{16}{13}=\frac{12}{13}$

$P(E_1/A)=\frac{12}{13}$

37. Let, $E_1$ :event that the student in hostel

$E_2$ : event that the student is day scholar

$A$ : be the event of the chosen student get grade $A$

$P(E_1)=60\mathrm{\%}=\frac{60}{100}=0.6$

$P(E_2)=40\mathrm{\%}=\frac{40}{100}=0.4$

$P(A/E_1)=P$ (student getting an $A$ grade is hostler) $=30\mathrm{\%}$ $=\frac{30}{100}=0.3$

$P(A/E_2)=P$ ( student getting an $A$ grade is a day scholar) $=20\mathrm{\%}=\frac{20}{100}=0.2$

Then, by Baye’s theorem,

$P(E_1/A)=\frac{P(E_1).P(A/E_1)}{P(E_1).P(A/E_1)+P(E_2).P(A/E_2)}$

$=\frac{0.6\times 0.3}{0.6\times 0.3+0.4\times 0.2}=\frac{0.18}{0.26}=\frac{18}{26}=\frac{9}{13}$

The candidates who have qualified PU B.A. LLB or B.Com.LL.B. (Hons.) 5-Years Integrated Course Entrance Test-2025 and have passed 10+2 examination are eligible for admission to the courses for the session 2025.

Yes, there are many private Design colleges in Delhi NCR. Some of them are mentioned below along with their tuition fees:

Disclaimer: This info. is taken from the official website/ sanctioning body and may vary.

36. Let $E_1$ be the event of selecting first bag

$E_2$ be the event of selecting second bag

$\therefore P(E_1)=P(E_2)=\frac{1}{2}$

Let $A$ be the event of getting a red ball

$P(A/E_1)=P$ (drawing a red ball from first bag)

$=\frac{4}{8}=\frac{1}{2}$

$P(A/E_2)=P$ (drawing a red ball from second bag)

$=\frac{2}{8}=\frac{1}{4}$

Then, probability of drawing a ball from the first bag which is red, is given by Baye’s theorem,

$P(E_1/A)=\frac{P(E_1).P(A/E_1)}{P(E_1).P(A/E_1)+P(E_2).P(A/E_2)}$

$=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\times \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\times \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\times \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}$

$=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}=\frac{2}{3}$