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39. Given, P (not E or not F) = 0.25

P (E'∪F') = 0.25

P (E∩F)' = 0.25

1 – P (E∩F) = 0.25

P (E∩F) = 1 – 0.25

P (E∩F) = 0.75≠ 0

Hence, E and F are not mutually exclusive events.

38. Given, P (E) = $\frac{1}{4}$

P (F) = $\frac{1}{2}$

P (E and F) = P (E∩F) = $\frac{1}{8}$

(i) P (E or F) = P (E∪F) = P (E) + P (F) – P (E∩F)

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

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

$=\frac{5}{8}$

(ii) P (not E and not F) = P (E'∩F') = P (E∪F)' = 1 – P (E∪F)

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

$=\frac{3}{8}$

13. For every four-digit number we have to count the permutation of 10 digits namely 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9 taken 4 at a time

However, these permutation will include those where 0 is at 1000's place.

So, fixing 0 at 1000's place and rearranging the remaining 9 digits taking 3 at a time.

Therefore, The required number = ¹⁰P₄ – ⁹P₃

= 5040 – 504

= 4536 ways

Some of the key indicators while ranking the universities include employment outcomes, international research network, and sustainability. The ranking criteria and their respective weightage are given in the table below:

37. 

Given P (A) = $\frac{3}{5}$

P (B) = $\frac{1}{5}$

As A and B are mutually exclusive events,

P (A∩B) = 0

Hence, P (A∪B) = P (A) + P (B) – P (A∩B)

$=\frac{3}{5}+\frac{1}{5}-0$

$=\frac{4}{5}$

No, Princeton University provides a test option for standardized test scores. It is not mandatory to submit the scores, however students prefer to submit them. Students must aim to submit ACT score between 33 and 36. Applicants who are submitting SAT, must aim between 1500 and 1580.

Based on the available data, securing a rank of 2585 in the Andhra Pradesh ICET 2025 exam may provide a reasonable chance of admission to the MCA self-financing programme at Sri Venkateswara University (SVU), Tirupati. For instance, in 2024, the Round 1 cutoff for the MCA programme was 22,418 for the General AI category . While the exact cutoff for 2025 is yet to be announced, your rank appears competitive for the self-financing quota

According to the admitted students' profiles, the student who scored 3.9 or above on GPA was enrolled in the program. Although the university provides test options for test scores, students still choose to submit them.

Students need to aim score between 33 and 36 for ACT, and between 1500 and 1580 for SAT. Student who manage to score near the ideal scores have high selection probability.

According to an unofficial source, the Princeton University acceptance rate is 4%. The low percentage indicates how difficult it is to get into the university and how strict it is. The severity of the secondary school record, class rank, academic GPA, standardized test scores, application essay, references, extracurricular activities, and personal traits are some of the important elements that the university takes into account. Students must take part in extracurricular activities, along with performing excellent in academics.

36. (i) Given P (A) = $\frac{1}{3}$

P (B) = $\frac{1}{5}$

P (A∩B) = $\frac{1}{15}$

So, P (A∪B) = P (A) + P (B) – P (A∩B)

$=\frac{1}{3}+\frac{1}{5}-\frac{1}{15}$

$=\frac{5+3-1}{15}$

P (A∪B) = $\frac{7}{15}$

(ii) Given P (A) = 0.35

P (B) =?

P (A∩B) = 0.25

P (A∪B) = 0.6

So, P (A∪B) = P (A) + P (B) – P (A∩B)

0.6 = 0.35 + P (B) – 0.25

P (B) = 0.6 – 0.35 + 0.25

P (B) = 0.5

(iii) Given P (A) = 0.5

P (B) = 0.35

P (A∩B) =?

P (A∪B) = 0.7

So, P (A∪B) = P (A) + P (B) – P (A∩B)

0.7 = 0.5 + 0.35 – P (A∩B)

P (A∩B) = 0.5 + 0.35 – 0.7

P (A∩B) = 0.15

Standardized test scores like ACT and SAT for undergraduate programs and GRE and GMAT for post graduate programs, are in general not mandatory for admission in Deakin University. But some programs might require these scores as programme specific requirements. The University focuses more on academic and English language requirements. But, Applicants can submit these test scores to strengthen their profile and increase the chances of their acceptance.

The total fee for a B.Tech in Computer Science Engineering (CSE) varies a lot depending on the college and its location.

• In government colleges, fees usually range between INR 1 lakh to INR 3 lakh for the entire 4-year course.

• In private colleges, it can be higher, anywhere from INR 3 lakh to INR 10 lakh or more for the full program, depending on the institute’s reputation and facilities.

35. Given P (A) = 0.5

P (B) = 0.7

And P (A∩B) = 0.6

As P (A∩B) > P (A) which is not possible.

The given probabilities are not consistently defined.

(ii) Given, P (A) = 0.5

P (B) = 0.4

And P (A∪B) = 0.8

So, P (A∪B) = P (A) + P (B) – P (A∩B)

0.8 = 0.5 + 0.4 – P (A∩B)

P (A∩B) = 0.5 + 0.4 – 0.8

P (A∩B) = 0.1

Hence, P (A∩B) < P (A) and P (A∩B) < P (B)

The given probabilities are consistently defined.

The Arena Animation has a placement cell that ensures placement assistance to every student. The college has released the 2025 placement data. Check the complete insights of Arena Animation South Ex placement from the table below:

In the last semester, candidates are prepared more for the real job. Hence, there are less number of subjects and more of practical knowledge involved. ICRI Pune MBA candidates also have to submit a dissertation in the last semester. Mentioned below are the subjects covered in the last semester of the programme:

• Airport Ramp Operations
• International Business
• Airport Terminal Operations

34. 

. Since 6 numbers are to be choosen as fixed from a set a given 20 number, the sample space is

$n\left(\right)=20C_6=\frac{20!}{6!\left(20-6\right)!}=\frac{20\times 19\times 18\times 17\times 16\times 15\times 14!}{2\times 3\times 4\times 5\times 6\times 14!}$

$=\frac{38760}{1}\times \frac{14!}{14!}=38760$

Let A: person wins the prize.

In order to win the prize the 6 number has to be correct i.e. all 6 of the number are to be choosen from fixed 6 numbers we have,

$n(S){=}^6{}_{\mathrm{C6}}=1$

? P (A) = $\frac{1}{38760}$ .

33.  The sample space of word is

S = {A, S, A, S, I, N, A, T, I, O, N}

So, n (S) = 13.

(i) Let A: word is a vowel

A = {A, I, A, I, O}

So, n (A) = 6

? P (A) = $\frac{6}{13}$

(ii) Let B: Word is a consonant

B = {S, N, T, N}

So, n (B) = 7

? P (B) = $\frac{7}{13}$ .

Here is the list showing various specialisations offered by a number of Fashion Design colleges in Chennai:

32. Let A be the event

Given that, P (A) = $\frac{2}{11}$

So, P (not A) = P (S) – P (A) = $1-\frac{2}{11}=\frac{11-2}{11}=\frac{9}{11}$