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At the time of filling out the application form for the MBA course at ICFAI University Raipur online, applicants are required to enclose self-attested copies of the following documents:

• Date of Birth CertificateClass 10 certificate
• Marks sheets of Class 12/ Diploma/ Graduation Degree. 
• Domicile Certificate in case of student belonging to the State of Chhattisgarh
• Nationality Certificate/Proof, for foreign students only. 
• Other certificates pertaining to academic and extracurricular activities.

48. If $A\subset B,$ then $A\cap B=A$

$P(A\cap B)=P(A)$ also $P(A)<P(B)$

Consider,

$P(A/B)=\frac{P(A\cap B)}{P(B)}=\frac{P(A)}{P(B)}\ne \frac{P(B)}{P(A)}$

$\Rightarrow P(A/B)=\frac{P(A\cap B)}{P(B)}=\frac{P(A)}{P(B)}$ $\_\_\_\_\_(1)$

We know, $P(B)\le 1$ $\Rightarrow \frac{1}{P(B)}\ge 1$

From eq. $(1)$ , we have

$\frac{P(A)}{P(B)}\ge P(A)$

$\Rightarrow P(A/B)\ge P(A)$ $\_\_\_\_\_(2)$

$\Rightarrow P(A/B)$ is not less than $P(A)$

Hence, from eq. $(2)$ it can be concluded that the relation given in alternative $(C)$ is correct.

47. Let, $E_1$ and $E_2$ be the event such that

$E_1:$ $A$ speak truth

$E_2:$ $A$ speak false

$X:$ that head appears

$P(E_1)=\frac{4}{5}$ and $P(E_2)=1-P(E_1)$ = $=1-\frac{4}{5}=\frac{1}{5}$

If a coin tossed, then it may result either head $(H)$ or tail $(T)$ . The probability of getting a head is $\frac{1}{2}$ whether $A$ speak truth or not.

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

The probability that there are actually a head is given by $P(E_1/X)$

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

$=\frac{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\times \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\times \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.+\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\times \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$=\frac{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\times \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\frac{1}{2}\left(\frac{4}{5}+\frac{1}{5}\right)}=\frac{4}{5}$

Option $(A)$ is correct $\frac{4}{5}$

The university has a total of 40 seats for the DPSRU BPharm in Ayurveda course. Aspirants are admitted as per the sanctioned seat intake. There are reserved seats as well as per the guidelines of the state government. It must be noted that the seat count mentioned here is as per the official website/ sanctioning body. It is subject to changes and hence, is indicative.

The JEE Advanced answer key is officially released by the IIT conducting the exam on the JEE Advanced website. Candidates can download it by logging in with their credentials. Along with the provisional key, authorities open a window to raise objections before publishing the final key, which is used for result preparation.

Candidates must appear for MAH CET exam to get admission to PGDM course. In connection to it, the institute has released the SIES College of Management Studies cutoff 2024 for the General AI category candidates. As per the MAH CET cutoff 2024, the percentile released for CMAT, XAT, MAT, and CAT stood at 80, 70, 80, and 25 for the General AI category candidates.

Minimum NEET score requirement may vary for each university, but most medical institutes in Russia consider the below NEET score requirements as the bare minimum to be eliigible for admissions:

• 50th percentile for students belonging to the General/EWS category
• 45th percentile for PwD category students 
• 40th percentile for SC/ST/OBC category students

45. Let, $E_1:$ event of time consume by $A$

$E_2:$ event of time consume by $B$

$E_3:$ event of time consume by $C$

$P(E_1)=50\mathrm{\%}=\frac{50}{100}=\frac{1}{2}$

$P(E_2)=30\mathrm{\%}=\frac{30}{100}=\frac{3}{10}$

$P(E_3)=20\mathrm{\%}=\frac{20}{100}=\frac{1}{5}$

Let, $A:$ event of producing the defective item. Therefore,

$P(A/E_1)=1\mathrm{\%}=\frac{1}{100}$

$P(A/E_2)=5\mathrm{\%}=\frac{5}{100}$

$P(A/E_3)=7\mathrm{\%}=\frac{7}{100}$

Therefore, by Baye’s theorem,

$P(E_1/A)=$ probability that the defective item was produced by $A$ ,

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

$=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\times \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$100$}\right.}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\times \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$100$}\right.+\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$10$}\right.\times \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$100$}\right.+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\times \raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$100$}\right.}$

$=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\times \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$100$}\right.}{\frac{1}{100}\left(\frac{1}{2}+\frac{15}{10}+\frac{7}{5}\right)}=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\times \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$100$}\right.}{\frac{1}{100}\left(\frac{5+15+14}{10}\right)}=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\raisebox{1ex}{$34$}\!\left/ \!\raisebox{-1ex}{$10$}\right.}$

$=\frac{1}{2}\times \frac{10}{34}=\frac{5}{34}$

44. Let, $E_1:$ event that outcome on the die is $5$ or $6$

$E_2:$ event that outcome on the die is $1,2,3$ or $4$

$A:$ event of getting exactly one head

$P(E_1)=\frac{2}{6}=\frac{1}{3}$

$P(E_2)=\frac{4}{6}=\frac{2}{3}$

$P(A/E_1)=$ probability of getting exactly one head by tossing the coin three times if she gets $5$ or $6$ $=\frac{3}{8}$

$P(A/E_2)=$ probabilit7y of getting exactly one head by tossing the coin three times if she gets $1,2,3$ or $4$ $=\frac{1}{2}$

Therefore, by Baye’s theorem,

$P(E_2/A)=$ probability that the girl threw $1,2,3$ or $4$ with die, if she obtained exactly one head,

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

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

$=\frac{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$24$}\right.+\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\frac{3+8}{24}}$

$=\frac{1}{3}\times \frac{24}{10}=\frac{8}{11}$

43. Let, $E_1:$ event that first group will win

$E_2:$ event that second group will win

$A:$ event that new product will produce

$P(E_1)=0.6=\frac{6}{10}=\frac{3}{5}$

$P(E_2)=0.4=\frac{4}{10}=\frac{2}{5}$

$P(A/E_1)=P$ (introducing new product by group $A$ ) $=0.7=\frac{7}{10}$

$P(A/E_2)=P$ ( introducing new product by group $B$ ) $=0.3=\frac{3}{10}$

Therefore, by Baye’s theorem,

$P(E_2/A)=$ probability that new product introduced was produced by second group

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

$=\frac{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\times \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$10$}\right.}{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$7$}\right.\times \raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$10$}\right.+\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\times \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$10$}\right.}$

$=\frac{\raisebox{1ex}{$6$}\!\left/ \!\raisebox{-1ex}{$50$}\right.}{\raisebox{1ex}{$21$}\!\left/ \!\raisebox{-1ex}{$50$}\right.+\raisebox{1ex}{$6$}\!\left/ \!\raisebox{-1ex}{$50$}\right.}$

$=\frac{6}{50}\times \frac{50}{27}=\frac{2}{9}$

The Easwari School of Liberal Arts at SRM University-AP is an integral part of SRM University, Andhra Pradesh, established in 2024. The institute is recognised by the University Grants Commission (UGC) and holds membership in the Association of Indian Universities (AIU). SRM University-AP is also accredited by the NAAC.

The Easwari School of Liberal Arts distinguishes itself through an innovative curriculum, a multidisciplinary educational approach, and a dynamic academic environment. The institute also offers a flexible academic structure and incorporates contemporary educational technologies to enrich the learning experience.

42. Let, $E_1:$ event items produced by $A$

$E_2:$ event items produced by $B$

$X:$ event that produced item was found to be defective

$P(E_1)=60\mathrm{\%}=\frac{3}{5}$

$P(E_2)=40\mathrm{\%}=\frac{2}{5}$

$P(X/E_1)=P$ (items produced by machine $A$ which is defective) $=2\mathrm{\%}=\frac{2}{100}$

$P(X/E_2)=P$ ( items produced by machine $B$ which is defective) $=1\mathrm{\%}=\frac{1}{100}$

Therefore, by Baye’s theorem,

$P(E_2/X)=$ probability that the randomly selected item was from machine $B$ ,which is defective,

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

$=\frac{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\times \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$100$}\right.}{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\times \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$100$}\right.+\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\times \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$100$}\right.}$

$=\frac{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$500$}\right.}{\raisebox{1ex}{$6$}\!\left/ \!\raisebox{-1ex}{$500$}\right.+\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$500$}\right.}$

$=\frac{2}{500}\times \frac{500}{8}=\frac{1}{4}$

41. Let, $E_1:$ event that the driver is scooter driver

$E_2:$ event that the driver is car driver

$E_3:$ event that the driver is truck driver

$A:$ event the person meets with accident

Total number of drivers $=2000+4000+6000=12000$

$P(E_1)=P$ (driver is a scooter driver) $=\frac{2000}{12000}=\frac{1}{6}$

$P(E_2)=P$ (driver is a car driver) $=\frac{4000}{12000}=\frac{1}{3}$

$P(E_3)=P$ (driver is a truck driver) $=\frac{6000}{12000}=\frac{1}{2}$

$P(A/E_1)=P$ (scooter driver met with an accident) $=0.01=\frac{1}{100}$

$P(A/E_2)=P$ (car driver met with an accident) $=0.03=\frac{3}{100}$

$P(A/E_3)=P$ (truck driver met with an accident) $=0.15=\frac{15}{100}$

The probability that the driver is scooter driver, given that he met with an accident is given by $P(E_1/A)$

By Baye’s theorem,

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