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HPBOSE 12th Sample papers are the question paper released by the Board at the beginning of the academic session. It shares the questions in a pattern similar to what you should expect in the main exam. They are provided to you to download and practice for better preparation.  However, HPBOSE 12th board exam question paper is the question paper you receive on the day of the exam. You need to go through the questions and answer them in the answer sheet provided by the board. You cannot know or practice the questions on the HPBOSE 12th board question paper prior to the exam.

Yes, Kolhan University at Chaibasa is a recognised university that is recognised by the University Grants Commission (UGC). All the courses are offered with a large number of specializations and are approved by various affiliating bodies. These affiliating bodies include prominent names like AICTE, NCTE, BCI, among several others. KU Chaibasa has an extensive network of affiliated and constituent colleges, with 21 constituent and 25+ affiliated colleges.

75. Let X represent the number of winning prizes in 50 lotteries. The trials are Bernoulli trials.

Clearly, X has a binomial distribution with n = 50 and p = 1/100

$\therefore q=1-p=1-\frac{1}{100}=\frac{99}{100}\\ \therefore P\left(X=x\right)={\mathrm{nC}}_x$
$q^{n-x}{p}^x={\mathrm{50C}}_x$
$\begin{array}{l}\mathrm{}{\left(\frac{99}{100}\right)}^{50-x}.{\left(\frac{1}{100}\right)}^x\end{array}$

P (winning at least once) $=P\left(X\ge 1\right)$

$=1-P\left(X<1\right)\\ =1-P\left(X=1\right)\\ =1-{\mathrm{50C}}_0$

$\begin{array}{l}\mathrm{}{\left(\frac{99}{100}\right)}^{50}\\ =1-1.{\left(\frac{99}{100}\right)}^{50}\\ =1-{\left(\frac{99}{100}\right)}^{50}\end{array}$

P (winning exactly once) $=P\left(X=1\right)$

$={\mathrm{50C}}_1$

$\begin{array}{l}\mathrm{}{\left(\frac{99}{100}\right)}^{49}.{\left(\frac{1}{100}\right)}^1\\ =50\left(\frac{1}{100}\right){\left(\frac{99}{100}\right)}^{49}\\ =\frac{1}{2}{\left(\frac{99}{100}\right)}^{49}\end{array}$

P (at least twice) $=P\left(X\ge 2\right)$

$\begin{array}{l}=1-P\left(X<2\right)\\ =1-P\left(X\le 1\right)\\ =1-\left[P\left(X=0\right)+P\left(X=1\right)\right]\\ =\left[1-P\left(X=0\right)\right]-P\left(X=1\right)\\ =1-{\left(\frac{99}{100}\right)}^{50}-\frac{1}{2}.{\left(\frac{99}{100}\right)}^{49}\\ =1-{\left(\frac{99}{100}\right)}^{49}.\left[\frac{99}{100}+\frac{1}{2}\right]\\ =1-{\left(\frac{99}{100}\right)}^{49}.\left(\frac{149}{100}\right)\\ =1-\left(\frac{149}{100}\right){\left(\frac{99}{100}\right)}^{49}\end{array}$

The final step of NIILM University BCom admission process is fee payment. Once selected, candidates must pay the tuition fee of INR 93,600 within the specified deadline to confirm their admission. It's important to check the official website for the latest fee and payment instructions.

Kolhan University is a UGC-recognized university based in Chaibasa, Jharkhand. The university provides a large number of courses to students with a variety of specializations. These courses are approved by various affiliating bodies like BCI, AICTE, NCTE, and several others. KU Chaibasa consists of 21 constituent colleges and over 25 affiliated colleges. The infrastructure of Kolhan University is designed keeping the students' holistic development in mind.

74. The repeated guessing of correct answers from multiple choice questions are Bernoulli trials. Let X represent the number of correct answers by guessing in the set of 5 multiple choice questions.

Probability of getting a correct answer is, p = 1/3

$\therefore q=1-p=1-\frac{1}{3}=\frac{2}{3}$

Clearly, X has a binomial distribution with n=5 and $P=\frac{1}{3}$

$\therefore P=\left(X=x\right)={\mathrm{nC}}_x$
$q^{n-x}{p}^x\\ ={\mathrm{5C}}_x$
$\begin{array}{l}\mathrm{}{\left(\frac{2}{3}\right)}^{5-x}.{\left(\frac{1}{3}\right)}^x\end{array}$

P (guessing more than 4 correct answers) $=P\left(X\ge 4\right)$

$=P\left(X=4\right)+P\left(X=5\right)\\ ={\mathrm{5C}}_4$
$\mathrm{}\left(\frac{2}{3}\right).{\left(\frac{1}{3}\right)}^4+{\mathrm{5C}}_5$
$\begin{array}{l}\mathrm{}{\left(\frac{1}{3}\right)}^5\\ =5.\frac{2}{3}.\frac{1}{81}+1.\frac{1}{243}\\ =\frac{10}{243}+\frac{1}{243}\\ =\frac{11}{243}\end{array}$

Students must ensure to solve atleast one previous year HPBOSE 12th question paper of Maths every day to efficiently prepare for the HPBOSE 12th Mathematics board exam. Make sure to practice the question papers during the last month of the preparation. 

NIILM University admission is primarily merit-based, focusing on academic performance in Class 12 exams from recognised boards. The university may conduct document verification or counselling if required.

According to latest data given by MEA of India, estimated no. of Indian students pursuing higher education in Singapore is 10,000.

Further, as per data published by India Today, the number of Indian student enrollment has seen a surge of about 25%, year-on-year, thanks to its lineant immigration policies, which makes it much easier for Indian students to get into the country, as compared to that of the US, UK & Canada.

The latest cutoff details for the B.Ed. program at Trinity Teacher's Training College are not publicly available. However, as per NCTE norms, the minimum eligibility requires a graduation degree with at least 50% marks, while candidates with qualifications like B.Tech. in Science/Mathematics need 55%. Admission is generally based on merit prepared from qualifying exam scores.

Applicants seeking admission to the NIILM BCom programme must submit a complete set of documents as per the university’s official requirements. These include the Class 12 mark sheet, proof of age (such as a birth certificate or Class 10 certificate), passport-sized photographs, and a valid category certificate (if applicable). Additional documents, if required, will be specified in the NIILM University BCom admission guidelines.

Yes, ICFAI University does offer scholarships to MBA students on the basis of their scores achieved in the last qualifying examination, and the university also offers scholarships to MBA students who perform decently in their MBA course in each semester. The university offers scholarships to MBA students ranging from INR 69,000 to INR 8,600. To know the proper details about the scholarships, candidates can refer to the student book available on the official website of the university.

73.  X is the random variable whose binomial distribution is B(6,1/2).

Therefore, n = 6 and p = ½

$\therefore q=1-p=1-\frac{1}{2}=\frac{1}{2}\\ Then,P\left(X=x\right)={\mathrm{nC}}_x$
$q^{n-x}{p}^x\\ ={\mathrm{6C}}_x$
$\mathrm{}{\left(\frac{1}{2}\right)}^{6-x}.{\left(\frac{1}{2}\right)}^x\\ ={\mathrm{6C}}_x$
$\begin{array}{l}\mathrm{}{\left(\frac{1}{2}\right)}^6\end{array}$

It can be seen that P (X=x) will be maximum, if ${\mathrm{6C}}_x$
 will be maximum.

$Then,{\mathrm{6C}}_0$
$\mathrm{}={\mathrm{6C}}_6$
$\mathrm{}=\frac{6!}{0!6!}=1\\ {\mathrm{6C}}_1$
$\mathrm{}={\mathrm{6C}}_5$
$\mathrm{}=\frac{6!}{1!5!}=6\\ {\mathrm{6C}}_2$
$\mathrm{}={\mathrm{6C}}_4$
$\mathrm{}=\frac{6!}{2!4!}=15\\ {\mathrm{6C}}_3$
$\begin{array}{l}\mathrm{}=\frac{6!}{3!3!}=20\end{array}$

The value of ${\mathrm{6C}}_3$
 is maximum. Therefore, for x=3, P(X=x) is maximum.

Therefore, P(X=3) is maximum.

72. Let X represent the number of correctly answered questions out of 20 questions.

The repeated tosses of a coin are Bernoulli trails. Since “head” on a coin represents the true answer and “tail” represents the false answer, the correctly answered questions are Bernoulli trials.

$\begin{array}{l}\therefore P=\frac{1}{2}\\ \therefore q=1-p=1-\frac{1}{2}=\frac{1}{2}\end{array}$

X has a binomial distribution with n=20 and $P=\frac{1}{2}$

$\therefore P=\left(X=x\right)={\mathrm{nC}}_x$
$q^{n-x}.p^x,$ where $x=0,1,2,...n$

$={\mathrm{20C}}_x$

$\mathrm{}{\left(\frac{1}{2}\right)}^{20-x}.{\left(\frac{1}{2}\right)}^x\\ ={\mathrm{20C}}_x$
$\begin{array}{l}\mathrm{}{\left(\frac{1}{2}\right)}^{20}\end{array}$

P (at least 12 questions answered correctly) $=P\left(X\ge 12\right)$

$=P\left(X=12\right)+P\left(X=13\right)+...+P\left(X=20\right)\\ ={\mathrm{20C}}_{12}$

$\mathrm{}{\left(\frac{1}{2}\right)}^{20}+{\mathrm{20C}}_{13}$
$\mathrm{}{\left(\frac{1}{2}\right)}^{20}+...+{\mathrm{20C}}_{20}$
$\mathrm{}{\left(\frac{1}{2}\right)}^{20}\\ ={\left(\frac{1}{2}\right)}^{20}.[20C_{12}$
$\mathrm{}+{\mathrm{20C}}_{13}$
$\mathrm{}+...+{\mathrm{20C}}_{20}$
$\begin{array}{l}\mathrm{}]\end{array}$

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71. Let X denote the number of balls marked with the digit 0 among the 4 balls drawn.

Since the balls are drawn with replacement, the trials are Bernoulli trials.

X has a binomial distribution with n = 4 and p = 1/10

$\therefore q=1-p=1-\frac{1}{10}=\frac{9}{10}\\ \therefore P\left(X=x\right)={\mathrm{nC}}_x$
$q^{n-x}.p^x,x=1,2,...n\\ ={\mathrm{4C}}_x$
$\begin{array}{l}\mathrm{}{\left(\frac{9}{10}\right)}^{4-x}.{\left(\frac{1}{10}\right)}^x\end{array}$

P (none marked with 0)=P (X=0)

$={\mathrm{4C}}_0$
$\begin{array}{l}\mathrm{}{\left(\frac{9}{10}\right)}^4.{\left(\frac{1}{10}\right)}^0\\ =1.{\left(\frac{9}{10}\right)}^4\\ ={\left(\frac{9}{10}\right)}^4\end{array}$

Andhra Medical College cutoff 2024 for the Round 1 NEET SS has been released for admission to PG courses at the college. As per the NEET SS cutoff data, the cutoff rank ranged between 629 and 1996 for admission for the General All India category students. For more cutoff related information on Andhra Medical College, click here. 

70. Let X represent the number of bulbs that will fuse after 150 days of use in an experiment of 5 trials. The trials are Bernoulli trials.

It is given that, p = 0.05

$\therefore q=1-p=1-0.05=0.95$

X has a binomial distribution with n=5 and $p=0.05$

$={\mathrm{5C}}_x$
$\mathrm{}{\left(0.95\right)}^{5-x}.{\left(0.05\right)}^x$

$P\left(none\right)=P\left(X=0\right)$

$={\mathrm{5C}}_0$
$\begin{array}{l}\mathrm{}{\left(0.95\right)}^5.{\left(0.05\right)}^0\\ =1\times {\left(0.95\right)}^5\\ ={\left(0.95\right)}^5\end{array}$

(ii) P (not more than one) $=P\left(X\le 1\right)$

$=P\left(X=0\right)+P\left(X=1\right)\\ =5C_0$
$\mathrm{}{\left(0.95\right)}^5\times {\left(0.05\right)}^0+{\mathrm{5C}}_1$
$\begin{array}{l}\mathrm{}{\left(0.95\right)}^4\times {\left(0.05\right)}^1\\ =1\times {\left(0.95\right)}^5+5\times {\left(0.95\right)}^4\times {\left(0.05\right)}^1\\ ={\left(0.95\right)}^5+\left(0.05\right){\left(0.95\right)}^4\\ ={\left(0.95\right)}^4\times 1.2\end{array}$

(iii) P (more than 1) $=P\left(X>1\right)$

$=1-P\left(X\le 1\right)$

$=1-P$ (not more than 1)

$=1-{\left(0.95\right)}^4\times 1.2$

(iv) P (at least one) $=P\left(X\ge 1\right)$

$=1-p\left(X<1\right)\\ =1-p\left(X=0\right)\\ =1-{\mathrm{5C}}_0$
$\begin{array}{l}\mathrm{}{\left(0.95\right)}^5\times {\left(0.05\right)}^0\\ =1-1\times {\left(0.95\right)}^5\\ =1-{\left(0.95\right)}^5\end{array}$