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9 months ago

International Institute of Information Technology, Pune International Institute of Information Technology, Pune Courses

Which courses are offered at I2IT Pune?

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Ishita B

Contributor-Level 10

I2IT Pune offers undergraduate courses to students. These courses are provided across the Engineering stream. At the UG level, the International Institute of Information Technology, Pune offers four-year full-time B.E. courses to students. B.E. is provided in three specialisations, namely Electronics & Telecommunication, Computer Engineering, and Information Technology.

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9 months ago

International Institute of Information Technology, Pune International Institute of Information Technology, Pune Eligibility

What is the admission criteria of I2IT, Pune?

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Aishwarya Bhatnagar

Contributor-Level 10

I2IT Pune is ranked 139th by Times 2024 for BTech. Admission to various courses is based on merit and entrance exam scores. The accepted entrance exams are MHT Main, MHT CET, and others. Candidates applying through JEE Main must first register for the JEE Main exam. Then appear for the JEE Main exam.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen Exemplar Solution

Without repetition of the numbers, four-digit numbers are formed with the numbers 0, 2, 3, 5. The probability of such a number divisible by 5 is:

(a) $\frac{1}{5}$

(b) $\frac{4}{5}$

(c) $\frac{1}{30}$

(d) $\frac{5}{9}$

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar $\begin{array}{l} F o u r d i g i t n u m b e r t h e d i g i t s 0, 2, 3, 5 w i t h o u t r e p e t i t i o n a n d d i v i s i b l e b y 5 w i t h t h e \\ g i v e n c o n d i t i o n i s \\ I f u n i t p l a c e b e f i l l e d w i t h 0 \\ T h e n t h e n u m b e r o f w a y s = 3 \times 2 \times 1 \times 1 = 6 \\ I f u n i t p l a c e b e f i l l e d w i t h 5 \\ T h e n t h e n u m b e r o f w a y s = 2 \times 2 \times 1 \times 1 = 4 \\ ∴ T o t a l n u m b e r o f w a y s = 6 + 4 = 10 \\ T o t a l n u m b e r o f w a y s o f a r r a n g i n g t h e d i g i t s 0, 2, 3, 5 t o f o r m 4 d i g i t n u m b e r s w i t h o u t \\ r e p e t i t i o n i s 3 \times 3 \times 2 \times 1 = 18 \\ ∴ Required p r o b a b i l i t y = \frac{10}{18} = \frac{5}{9} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen Exemplar Solution

Seven persons are to be seated in a row. The probability that two particular persons sit next to each other is:

(a) $\frac{1}{3}$

(b) $\frac{1}{6}$

(c) $\frac{2}{7}$

(d) $\frac{1}{2}$

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T h e t w o p a r t i c u l a r p e r s o n s t o b e s e a t e d n e x t e a c h o t h e r t h e n, t h e y f o r m o n e g r o u p . \\ N o w t h e p e r m u t a t i o n o f 6 p e r s o n s = 6! \times 2! \\ a n d T o t a l n u m b e r o f p e r m u t a t i o n o f 7 p e r s o n s = 7! \\ ∴ Required p r o b a b i l i t y = \frac{6! \times 2!}{7!} \\ = \frac{6! \times 2}{7 \times 6!} = \frac{2}{7} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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9 months ago

BPP University Colleges in UK

Is BPP University good for Engineering?

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen Exemplar Solution

While shuffling a pack of 52 playing cards, 2 are accidentally dropped. Find the probability that the missing cards are of different colors

(a) $\frac{29}{52}$

(b) $\frac{1}{2}$

(c) $\frac{26}{51}$

(d) $\frac{27}{51}$

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} W e k n o w t h a t o u t o f 52 p l a y i n g c a r d s 26 a r e o f r e d a n d 26 a r e o f b l a c k c o l o u r . \\ ∴ P (b o t h c a r d s o f d i f f e r e n t c o l o u r) = \frac{26}{52} \times \frac{26}{51} + \frac{26}{52} \times \frac{26}{51} \\ = 2 \times \frac{26}{52} \times \frac{26}{51} = \frac{26}{51} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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9 months ago

NIT Jalandhar - Dr. B. R. Ambedkar National Institute of Technology Affiliation

Does National Institute of Technology, Jalandhar has any collaboration with any college or university?

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Liyansha Jain

Contributor-Level 10

Yes, NITJ has collaborated with various colleges/universities. The college has signed MoUs with various industries and organisations such as UiT The Arctic University Norway, University of Manitoba, Canada, Technical Sciences? A? AK, University of Kragujevac, Arba Minch University, Arba Minch Ethiopia, and others.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen Exemplar Solution

Three numbers are chosen from 1 to 20. Find the probability that they are not consecutive:

(a) $\frac{186}{190}$

(b) $\frac{187}{190}$

(c) $\frac{188}{190}$

(d) 18/²⁰C₃

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} S e t o f t h r e e c o n s e c u t i v e n u m b e r s f r o m 1 t o 20 a r e 1, 2, 3; 2, 3, 4; 3, 4, 5; \dots; 18, 19, 20 . \\ S o, t h e p r o b a b i l i t y t h a t t h e n u m b e r s a r e consecutive \\ = \frac{18}{{}_{20}{C_{3}}} = \frac{18}{\frac{20!}{3! 7!}} = \frac{18.3! 7!}{20!} \\ = \frac{18 \times 3 \times 2 \times 17!}{20 \times 19 \times 18 \times 17!} = \frac{3 \times 2}{20 \times 19} = \frac{3}{190} \\ ∴ P (t h r e e n u m b e r s a r e n o t consecutive) = 1 - \frac{3}{190} = \frac{187}{190} \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen Exemplar Solution

In a non-leap year, the probability of having 53 Tuesdays or 53 Wednesdays is:

(a) $\frac{1}{7}$

(b) $\frac{2}{7}$

(c) $\frac{3}{7}$

(d) None of these

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Payal Gupta

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

$\begin{array}{l} T h e r e a r e 365 d a y s i n a l e a p y e a r a n d t h e r e a r e 7 d a y s i n a w e e k . \\ ∴ 365 \div 7 = 52 w e e k s + 1 d a y \\ S o, t h i s d a y m a y b e T u e s d a y o r W e d n e s d a y . \\ S o, t h e r e q u i r e d p r o b a b i l i t y = \frac{1}{7} . \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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9 months ago

Siddaganga Institute of Technology CG PET

I want to know about the latest information for CG PET Cut-off at Siddaganga Institute of Technology.

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9 months ago

Delhi University (DU) CUET Scores

Which college in DU can I get if I score 680 out of 1250 in CUET?

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9 months ago

NIT Jalandhar: Placements, Admission 2024, Cut off, Courses, Fees, Ranking NIT

Is NIT Jalandhar part of Top 10 NITs?

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Raj Kumar

Contributor-Level 10

No, NIT Jalandhar is not among the Top 10 NITs present in India. It is at the 12th position when compared with other NITs as per the NIRF report for 2025. It stands at the 55th position in the NIRF ranking 2025 under the Engineering category. Students can click here to check the complete details about the NIT Jalandhar's position among the other NITs.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen Exemplar Solution

Determine the probability p, for each of the following events.

(a) An odd number appears in a single toss of a fair die.

(b) At least one head appears in two tosses of a fair coin.

(c) A king, 9 of hearts, or 3 of spades appears in drawing a single card from a well shuffled ordinary deck of 52 cards.

(d) The sum of 6 appears in a single toss of a pair of fair dice.

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} (a) P o s s i b l e o u t c o m e o f a single t h r o w o f d i e \\ S = {1, 2, 3, 4, 5, 6} o u t o f w h i c h 1, 3, 5 a r e o d d \\ ∴ R e q u i r e d p r o b a b i l i t y = \frac{3}{6} = \frac{1}{2} \\ (b) W h e n a f a i r c o i n i s t o s s e d t w i c e, t h e n t h e s a m p l e s p a c e \\ S = {H H, H T, T H, T T} \\ ∴ P r o b a b i l i t y o f g e t t i n g a t l e a s t o n e h e a d (H H, H T, T H) = \frac{3}{4} \\ (c) F a v o u r a b l e e v e n t s a r e 4 k i n g s + 2 o f h e a r t s + 3 o f s p a d e s = 4 + 1 + 1 = 6 = \frac{6}{52} = \frac{3}{26} \\ (d) W h e n a p a i r o f d i c e i s r o l l e d, t h e n t o t a l n u m b e r o f s a m p l e s p a c e = 36 o u t o f w h i c h \\ (1, 5), (5, 1), (2, 4), (4, 2) a n d (3, 3) a r e t h e f a v o u r a b l e e v e n t s \\ ∴ R e q u i r e d p r o b a b i l i t y = \frac{5}{36} . \end{array}$

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9 months ago

NIT Jalandhar: Placements, Admission 2024, Cut off, Courses, Fees, Ranking NIT Jalandhar - Dr. B. R. Ambedkar National Institute of Technology Courses

Which courses are offered at NIT Jalandhar?

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Neha Pradhan

Contributor-Level 10

NIT Jalandhar offers UG, PG and PhD programmes in the fields of Management, Sciences, Engineering and Technology. The popular programmes of this institute are BTech and MTech. In addition, NIT Jalandhar also offers PG Diploma programmes in Management and Engineering. Below is the list of programmes along with the NITJ eligibility and fees:

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9 months ago

NIT Jalandhar - Dr. B. R. Ambedkar National Institute of Technology Eligibility NIT Jalandhar - Dr. B. R. Ambedkar National Institute of Technology Cut-off

How can I get admission to Jalandhar NIT?

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Manpreet Sharma

Contributor-Level 10

For admission to its M.Tech programme, NIT Jalandhar accepts the GATE scores. The institute accepts Joint Admission Test (IIT JAM) scores for admission to its MSc programme. Furthermore, the National Institute of Technology accepts national-level and state-level entrance exams such as CAT, MAT, and CMAT for admission to its MBA programme. After qualifying for the respective entrance exams, interested candidates must apply for the relevant counselling sessions, such as JoSAA and CCMT, to gain admission to various courses offered by the institute.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen Exemplar Solution

A sample space consists of 9 elementary outcomes $e_{1}, e_{2}, \dots, e_{9}$ whose probabilities are:

$P (e_{1}) = P (e_{2}) = 0.08$ , $P (e_{3}) = P (e_{4}) = P (e_{5}) = 0.1$ , $P (e_{6}) = P (e_{7}) = 0.2$ , $P (e_{8}) = P (e_{9}) = 0.07$ .

Suppose $A = {e_{1}, e_{5}, e_{8}}$ , $B = {e_{2}, e_{5}, e_{8}, e_{9}}$ .

(a) Calculate $P (A)$ , $P (B)$ , and $P (A \cap B)$ .

(b) Using the addition law of probability, calculate $P (A \cup B)$ .

(c) List the composition of the event $A \cup B$ , and calculate $P (A \cup B)$ by adding the probabilities of the elementary outcomes.

(d) Calculate $P (B^{'})$ from $P (B)$ , also calculate $P (B^{'})$ directly from the elementary outcomes of $B$ .

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} G i v e n t h a t : S = {e_{1}, e_{2}, e_{3}, e_{4}, e_{5}, e_{6}, e_{7}, e_{8}, e_{9}} \\ A = {e_{1}, e_{5}, e_{5}} a n d B = {e_{2}, e_{5}, e_{5}, e_{9}} \\ P (e_{1}) = P (e_{2}) = 0.08 \\ P (e_{3}) = P (e_{4}) = P (e_{5}) = 0.1 \\ P (e_{6}) = P (e_{7}) = 0.2 \\ P (e_{8}) = P (e_{9}) = 0.07 \\ (a) P (A) = P (e_{1}) + P (e_{5}) + P (e_{8}) \\ = 0.08 + 0.1 + 0.07 = 0.25 \\ (b) P (A \cup B) = P (A) + P (B) - P (A \cap B) \dots (i) \\ B u t P (B) = P (e_{2}) + P (e_{5}) + P (e_{8}) + P (e_{9}) \\ = 0.08 + 0.1 + 0.07 + 0.07 = 0.32 \\ a n d P (A \cap B) = {e_{5}, e_{8}} = P (e_{5}) + P (e_{8}) \\ = 0.1 + 0.07 = 0.17 \\ P u t t i n g t h e v a l u e s i n e q n . (i) w e g e t \\ P (A \cup B) = 0.25 + 0.32 - 0.17 = 0.40 \\ (c) P (A \cup B) = {e_{1}, e_{2}, e_{5}, e_{8}, e_{9}} \\ ∴ P (A \cup B) = P (e_{1}) + P (e_{2}) + P (e_{5}) + P (e_{8}) + P (e_{9}) \\ = 0.08 + 0.08 + 0.1 + 0.07 + 0.07 = 0.40 \\ (d) &t \end{array}$

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen Exemplar Solution

A card is drawn from a deck of 52 cards. Find the probability of getting a king or a heart or a red card.

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} T o t a l n u m b e r o f c a r d s = 52 \\ F a v o u r a b l e e v e n t s = 4 k i n g s + 13 h e a r t s + 26 r e d - 13 - 2 = 28 \\ ∴ R e q u i r e d p r o b a b i l i t y = \frac{28}{52} = \frac{7}{13} . \end{array}$

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9 months ago

NIT Jalandhar: Placements, Admission 2024, Cut off, Courses, Fees, Ranking NIT Jalandhar - Dr. B. R. Ambedkar National Institute of Technology Placement

Which companies generally recruit from Jalandhar NIT?

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Jagriti Pruthi

Contributor-Level 10

Many notable companies visited NIT Jalandhar campus placements and some of them are:

NIT Jalandhar Placement: Top Recruiters
Google	CVent
Microsoft	Amazon
Optum	Nvidia
Expedia.	Bajaj Finserv

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9 months ago

NIT Jalandhar: Placements, Admission 2024, Cut off, Courses, Fees, Ranking NIT Jalandhar - Dr. B. R. Ambedkar National Institute of Technology Courses

How can students apply for admission to various courses at NIT Jalandhar?

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Rashmi Arora

Contributor-Level 10

The mode of Dr B R Ambedkar National Institute of Technology Jalandhar course admissions is online. NIT Jalandhar course admissions are entrance-based. The institute offers the BTech programme at the undergraduate level and the MBA, MSc and M.Tech programmes at the PG level. For admission to its BTech programme, Dr BR Ambedkar National Institute of Technology Jalandhar accepts the JEE Main scores, followed by the Joint Seat Allocation Authority (JoSAA) counselling.

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9 months ago

Maths NCERT Exemplar Solutions Class 11th Chapter Sixteen Exemplar Solution

If the letters of the word ASSASSINATION are arranged at random, find the probability that:

(a) Four S’s come consecutively in the word.

(b) Two I’s and two N’s come together.

(c) All A’s are not coming together.

(d) No two A’s are coming together.

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Payal Gupta

Contributor-Level 10

This is a long answer type question as classified in NCERT Exemplar

$\begin{array}{l} T o t a l n u m b e r o f w o r d i s A S S A S S I N A T I O N a r e 13 . \\ W h e r e, w e h a v e 3 A' s, 4 S' s, 2 I' s, 2 N' s, 1 T' s a n d 1 O' s . \\ (a) I f 4 S' s c o m e consecutively i n t h e w o r d, t h e n a r r a n g e m e n t m a y b e a s f o l l o w s : \\ \underset{1 G r o u p}{S S S S} \underset{9 o t h e r s}{\underset{?}{A A A I I N N T O}} \\ ∴ N u m b e r o f w o r d s w h e n a l l S' s a r e t o g e t h e r = \frac{10!}{3! 2! 2!} \end{array}$

$\begin{array}{l} a n d t h e t o t a l n u m b e r o f w o r d f o r m e d f r o m t h e w o r d s A S S A S S I N A T I O N = \frac{13!}{3! 4! 2! 2!} \\ ∴ R e q u i r e d p r o b a b i l i t y = \frac{\frac{10!}{3! 2! 2!}}{\frac{13!}{3! 4! 2! 2!}} = \frac{10!}{3! 2! 2!} \times \frac{3! 4! 2! 2!}{13!} \\ = \frac{10! 4!}{13!} = \frac{10! \times 4 \times 3 \times 2}{13 \times 12 \times 11 \times 10!} = \frac{2}{143} \\ (b) I f 2 I' s a n d 2 N' s c o m e t o g e t h e r t h e n t h e r e a r e 10 a l p h a b e t s \\ N u m b e r o f w o r d s w h e n 2 I' s a n d 2 N' s c o m e t o g e t h e r = \frac{10!}{3! 4!} \times \frac{4!}{2! 2!} \\ ∴ R e q u i r e d p r o b a b i l i t y = \frac{\frac{10!}{3! 4!} \times \frac{4!}{2! 2!}}{\frac{13!}{3! 4! 2! 2!}} = \frac{10!}{3! 4!} \times \frac{4!}{2! 2!} \times \frac{3! 4! 2! 2!}{13!} \\ = \frac{4! 10!}{13!} = \frac{4 \times 3 \times 2 \times 10!}{13 \times 12 \times 11 \times 10!} = \frac{2}{143} \\ (c) I f a l l A' s a r e coming t o g e t h e r, t h e n t h e r e a r e 11 a l p h a b e t s \\ N u m b e r o f w o r d s w h e n a l l A' s c o m e t o g e t h e r = \frac{11!}{4! 2! 2!} \\ ∴ P r o b a b i l i t y w h e n a l l A' s c o m e t o g e t h e r = \frac{\frac{11!}{4! 2! 2!}}{\frac{13!}{4! 3! 2! 2!}} = \frac{11!}{4! 2! 2!} \times \frac{4! 3! 2! 2!}{13!} \\ = \frac{11! 3!}{13!} = \frac{6}{13 \times 12} = \frac{1}{26} \\ ∴ R e q u i r e d p r o b a b i l i t y w h e n a l l A' s d o n o t c o m e t o g e t h e r = 1 - \frac{1}{26} = \frac{25}{26} \\ (d) I f n o t w o A' s a r e t o g e t h e r, t h e n a r r a n g i n g t h e a l p h a b e t s e x c e p t A' s \\ - S - S - S - S - I - N - T - I - O - N - \\ N u m b e r o f w a y s o f a r r a n g i n g a l l a l p h a b e t s e x c e p t A' s = \frac{10!}{4! 2! 2!} \\ T h e r e a r e 11 v a c a n t p l a c e s b e t w e e n t h e s e a l p h a b e t s . \\ ∴ 3 A' s c a n b e p l a c e d i n 11 p l a c e s i n {}_{11}{C_{3}} w a y s = \frac{11!}{3! 8!} \\ ∴ T o t a l n u m b e r o f w o r d s w h e n n o t w o A' s c o m e t o g e t h e r = \frac{11!}{3! 8!} \times \frac{10!}{4! 2! 2!} \\ ∴ R e q u i r e d p r o b a b i l i t y = \frac{11!}{3! 8!} \times \frac{10!}{4! 2! 2!} \times \frac{4! 3! 2! 2!}{13!} \\ = \frac{10!}{8! \times 13 \times 12} = \frac{10 \times 9 \times 8!}{8! \times 13 \times 12} = \frac{10 \times 9}{13 \times 12} = \frac{15}{26} \end{array}$

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