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The seat intake for MPharm at Nehru College of Pharmacy is 45. Candidates are shortlisted on the basis of their performance in the GPAT. Candidates who clear the admission rounds receive the final admission offer. Candidates are required to pay the course fee to secure their seat. The seat intake can vary from specialisation to specialisation, candidates are required to go to the official website to learn more about the seat intake. 

You will need the details mentioned on your PSEB 10th admit card to check your result. Its okay if you have the photocopies of your hall ticket and lost the original one. 

However, it is always advised to keep the documents like Punjab board Class 10 admit card secured till the announcement of results.

This is a  Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Weknowthat:Var\left(X\right)=E\left(X^2\right)-{\left[E\left(X\right)\right]}^2\\ E\left(X\right)={\displaystyle \sum _{i=1}^n{p}_i}{x}_i\\ =0\times \frac{1}{6}+1\times \frac{5}{18}+2\times \frac{2}{9}+3\times \frac{1}{6}+4\times \frac{1}{9}+5\times \frac{1}{18}\\ =0+\frac{5}{18}+\frac{4}{9}+\frac{3}{6}+\frac{4}{9}+\frac{5}{18}=\frac{5+8+9+8+5}{18}=\frac{35}{18}\\ andE\left(X^2\right)=0\times \frac{1}{6}+1\times \frac{5}{18}+4\times \frac{2}{9}+9\times \frac{1}{6}+16\times \frac{1}{9}+25\times \frac{1}{18}\\ =\frac{5}{18}+\frac{8}{9}+\frac{9}{6}+\frac{16}{9}+\frac{25}{18}=\frac{5+16+27+32+25}{18}=\frac{105}{18}\\ \therefore Var\left(X\right)=\frac{105}{18}-\frac{35}{18}\times \frac{35}{18}=\frac{1890-1225}{324}=\frac{665}{324}\\ Hence,therequiredis\frac{665}{324}.\end{array}$

The PTE score remains valid for a period of 2 years after which you will be required to take another PTE exam. During these two years, one can share their PTE score report with as many universities as they want.

Clusters Academy of Excellence Graphic Design Certificate course syllabus includes various subjects. All these subjects deal with different domains of Design. Students can check out the below list to know some of the subjects taught in the course:

• Design Principles
• Software and Tools
• Typography
• Web and Mobile Design
• Design.2 Theory
• Branding and Identity
• Motion Graphics and Animation
•  Portfolio Development

This is a  Short Answer Type Questions as classified in NCERT Exemplar

Sol:

PGDM at Birla Institute of Management Technology is considered a good choice. The institute offers academic excellence, good infrastructure facilities and strong faculty. This in turn offers a number of opportunities for students such as strong placement records, decent salaries and a variety of opportunities. Students who are planing to study PGDM can definitely consider Birla Institute of Management Technology. 

For the comfort of study abroad aspirants, PLC Group – the conducting body of the PTE Academic exam had launched the PTE Academic at Home version. The test could be taken from the comfort of the homes of the candidates.

However, it is important to note that the PTE Academic online version of the test has now been discontinued by the Pearson PLC Group. Hence, currently students cannot take the PTE exam from home. 

The total fees for BTech at IIT Jammu is cumulative of multiple components. A few of these are tuition fees, hostel fees, security fee, and more. Students who get selected have to pay the first semester fee or the prescribed amount to confirm their candidature. In the following table, IIT Jammu BTech fee structure is showcased:

NOTE: The above-mentioned fee is taken from the official website/ sanctioning body. It is still subject to changes and hence, is indicative.

This is a  Short Answer Type Questions as classified in NCERT Exemplar

Sol:

This is a  Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Giventhat:S=\left\{1,2,3,\dots ,n\right\}\\ \therefore P\left(r\le p/s\le p\right)=\frac{P\left(P\cap S\right)}{P\left(S\right)}=\frac{p-1}{n}\times \frac{n}{n-1}=\frac{p-1}{n-1}\\ Hence,therequiredprobabilityis\frac{p-1}{n-1}.\end{array}$

This is a  Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Let{E}_1=TheeventthatapersonselectedisofbloodgroupO\\ E_2=Theeventthatapersonselectedisofothergroup\\ andH=Theeventthatselectedpersonislefthanded.\\ P\left(E_1\right)=0.30,P\left(E_2\right)=0.70,P\left(H/E_1\right)=0.06,P\left(H/E_2\right)=0.10\\ \therefore UsingBaye\text{'}sTheorm,weget\\ P\left(E_1/H\right)=\frac{P\left(E_1\right).P\left(H/E_1\right)}{P\left(E_1\right).P\left(H/E_1\right)+P\left(E_2\right).P\left(H/E_2\right)}\\ =\frac{0.30\times 0.06}{0.30\times 0.06+0.70\times 0.10}=\frac{0.018}{0.018+0.070}=\frac{0.018}{0.088}=\frac{9}{44}\\ Hence,therequiredprobabilityis\frac{9}{44}.\end{array}$

The deadline to fill out the BTech application form at Guru Nanak Dev University Regional Campus was June 10, 2025. Therefore, the deadline has been announced. Moreover, the counselling process by the university's end is also done. Now, candidates who receive admission letters can continue their courses further after paying an adequate amount of course fees.

This is a  Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Let{E}_1=Eventthatthecoinisfair\\ E_2=Eventthatthecoinis2-headed\\ andH=Eventthatthetossedcoingetshead.\\ P\left(E_1\right)=\frac{1}{2},P\left(E_2\right)=\frac{1}{2},P\left(H/E_1\right)=\frac{1}{2},P\left(H/E_2\right)=1\\ \therefore UsingBaye\text{'}sTheorm,weget\\ P\left(E_1/H\right)=\frac{P\left(E_1\right).P\left(H/E_1\right)}{P\left(E_1\right).P\left(H/E_1\right)+P\left(E_2\right).P\left(H/E_2\right)}\\ =\frac{\frac{1}{2}.\frac{1}{2}}{\frac{1}{2}.\frac{1}{2}+\frac{1}{2}.1}=\frac{\frac{1}{4}}{\frac{1}{4}+\frac{1}{2}}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}\\ Hence,therequiredprobabilityis\frac{1}{3}.\end{array}$

No, Cyber Media covers more than just social media. Cyber Media is the utilization of every digital medium that disseminates messages, entertains, or connects people. Streaming services like Netflix and Prime Video, news websites like Hindustan Times, Business Standard, etc., Video-sharing platforms like YouTube, everything is a part of Cyber Media.

This is a  Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l}LetXbetherandomdenotingabulbtobedefective.\\ Here,n=10,p=\frac{1}{50},q=1-\frac{1}{50}=\frac{49}{50}\\ WeknowthatP\left(X=r\right)={}_n{C}_r{p}^r{q}^{n-r}\\ \left(i\right)Noneofthebulbsisdefective,i.e.,r=0\\ P\left(x=0\right)={}_{10}{C}_0{\left(\frac{1}{50}\right)}^0{\left(\frac{49}{50}\right)}^{10-0}={\left(\frac{49}{50}\right)}^{10}\\ \left(ii\right)Exactlytwobulbsaredefective,i.e.,r=2\\ \therefore P\left(x=2\right)={}_{10}{C}_2{\left(\frac{1}{50}\right)}^2{\left(\frac{49}{50}\right)}^{10-2}=45.\frac{{\left(49\right)}^8}{{\left(50\right)}^{10}}=45\times {\left(\frac{1}{50}\right)}^{10}\times {\left(49\right)}^8\\ \left(iii\right)Morethan8bulbsworkproperly\\ Wecansaythatlessthan2bulbsaredefective\\ P\left(X<2\right)=P\left(x=0\right)+P\left(x=1\right)\\ ={}_{10}{C}_0{\left(\frac{1}{50}\right)}^0{\left(\frac{49}{50}\right)}^{10-0}+{}_{10}{C}_1{\left(\frac{1}{50}\right)}^1{\left(\frac{49}{50}\right)}^9={\left(\frac{49}{50}\right)}^{10}+\frac{1}{5}{\left(\frac{49}{50}\right)}^9\\ ={\left(\frac{49}{50}\right)}^9\left(\frac{49}{50}+\frac{1}{5}\right)={\left(\frac{49}{50}\right)}^9\left(\frac{59}{50}\right)=\frac{59{\left(49\right)}^9}{{\left(50\right)}^{10}}\end{array}$