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12. Let $B$ and $G$ denote boy and girl respectively.

Sample space, $S=\{GG,BB,GB,BG\}$

Let, $E:$ Both are girls

$E=\{G,G\}$

i. Let $,F:$ youngest is a girl

$F=\{GG,B,G\}$

$E\cap F=\{G,G\}$ $,P(E\cap F)=\frac{1}{4}$

$P(E/F)=\frac{P(E\cap F)}{P(F)}=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}=\frac{1}{2}$

ii. Let $A:$ at least one is girl

$A=\{GG,GB,BG\}$

$P(A)=\frac{3}{4}$

$E\cap A=\left\{GG\right\}$

$P(E\cap A)=\frac{1}{4}$

$P(E/A)=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}=\frac{1}{3}$

11. The sample space of given condition is

A $S=\{1,2,3,4,5,6\}$

$E=\{1,3,5\}$ $,F=\{2,3\}$ $,G=\{2,3,4,5\}$

$P(E)=\frac{3}{6}=\frac{1}{2}$ $,P(F)=\frac{2}{6}=\frac{1}{3}$ $P(G)=\frac{4}{6}=\frac{2}{3}$

$\mathrm{(i)\; P}(E/F)$ and $P(F/E)$

$P(E\cap F)=\frac{1}{6}$

$P(E/F)=\frac{P(E\cap F)}{P(F)}=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=\frac{1}{2}$

and $P(F/E)=\frac{P(F\cap E)}{P(E)}=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{1}{3}$

$\mathrm{(ii)\; P}(E/G)$ and $P(G/E)$

$P(E\cap G)=\frac{2}{6}=\frac{1}{3}$

$P(E/G)=\frac{P(E\cap G)}{P(G)}=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=\frac{1}{2}$

$P(G/E)=\frac{P(G\cap E)}{P(E)}=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\frac{2}{3}$

$\mathrm{(iii)\; P}((E\cup F)/G)$ and $P((E\cap F)/G)$

$(E\cup F)=\{1,2,3,5\}$

$(E\cup F)\cap G=\{1,2,3,5\}\cap \{2,3,4,5\}$

$=\{2,3,5\}$

and $E\cap F=\left\{3\right\}$

$(E\cap F)\cap G=\left\{3\right\}$

$P(E\cup G)=\frac{4}{6}=\frac{2}{3}$

Now,

$P((E\cup F)\cap G)=\frac{3}{6}=\frac{1}{2}$

So,

$P((E\cup F)/G)=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=\frac{3}{4}$

$P((E\cap F)/G)=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$6$}\right.}{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=\frac{3}{12}=\frac{1}{4}$

The SOP's full form is the Statement of Purpose (SOP). An SOP is an important document and an integral part of your application procedure for those looking to study abroad. A statement of purpose is written by the applicant on his/her own only. 

Yes, BSc is available at IHM Bangalore. This course is made available in affiliation with the National Council for Hotel Management and the Indira Gandhi National Open University. Check out the below table to know  the highlights of the programme:

Note: The above information has been sourced from various sources. Hence it is indicative.

To be considered for IHM Bangalore BSc admission, it is essential that candidates fulfil the eligibility conditions set by the institute. As per the eligibility criteria, candidates who have passed Class 12 with English as a subject of study are eligible to apply. Moreover, the candidates must also have a valid NHMCT JEE score.

10. When two die are rolled, the sample space has $6\times 6=36$

Let $A$ be obtaining a sum greater than

$9=\{(4,6),(5,5),(5,6),(6,4),(6,5),(6,6)\}$

And $B$ be black die result in $5$ ,

$B=\{(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)\}$

$A\cap B=\{(5,5),(5,6)\}$

$\mathrm{a.\; P}(A/E)=\frac{P(A\cap B)}{P(B)}=\frac{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$36$}\right.}{\raisebox{1ex}{$6$}\!\left/ \!\raisebox{-1ex}{$36$}\right.}=\frac{1}{3}$

b. Let,

$E:$ sum of observation is $8$

$=\{(2,6),(3,5),(4,4),(5,3),(6,2)\}$

$F:$ Red die resulted in number less than $4$

$=\left\{\begin{array}{l}(1,1),(1,2),(1,3),(2,1),(2,2),(2,3)\\ (3,1),(3,2),(3,3),(4,1),(4,2),(4,3)\\ (5,1),(5,2),(5,3),(6,1),(6,2),(6,3)\end{array}\right\}$

$E\cap F=\{(5,3),(6,2)\}$

$P(F)=\frac{18}{36}$ and $P(E\cap F)=\frac{2}{36}$

$P(E/F)=\frac{P(E\cap F)}{P(F)}=\frac{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$36$}\right.}{\raisebox{1ex}{$18$}\!\left/ \!\raisebox{-1ex}{$36$}\right.}=\frac{2}{18}=\frac{1}{9}$

9. The sample space of given condition is

$S=\{(MFS,MSF,FMS,FSM,SMF,SFM)\}$

$E:\{MFS,FMS,SMF,SFM\}$

$F:\{MFS,SFM\}$

$P(F)=\frac{2}{6}=\frac{1}{3}$ ,  $P(E\cap F)=\frac{2}{6}=\frac{1}{3}$

$P(E/F)=\frac{P(E\cap F)}{P(F)}=\frac{\frac{1}{3}}{\frac{1}{3}}=1$

If a die is thrown three times, then the number of elements in the sample space will be $6*6*6=216$

Here, $E=$ $\left\{\begin{array}{l}(1,1,4),(1,2,4),(1,3,4),(1,4,4),(1,5,4),(1,6,4)\\ (2,1,4),(2,2,4),(2,3,4),(2,4,4),(2,5,4),(2,6,4)\\ (3,1,4),(3,2,4),(3,3,4),(3,4,4),(3,5,4),(3,6,4)\\ (4,1,4),(4,2,4),(4,3,4),(4,4,4),(4,5,4),(4,6,4)\\ (5,1,4),(5,2,4),(5,3,4),(5,4,4),(5,5,4),(5,6,4)\\ (6,1,4),(6,2,4),(6,3,4),(6,4,4),(6,5,4),(6,6,4)\end{array}\right\}$

$F=\{(6,5,1),(6,5,2),(6,5,3),(6,5,4),(6,5,5),(6,5,6)\}$

$?E?F=\{(6,5,4)\}$

$i.e.,P(F)=\frac{6}{216}$ $,P(E?F)=\frac{1}{216}$

$P(E/F)=\frac{P(E?F)}{P(F)}=\frac{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$216$}\right.}{\raisebox{1ex}{$6$}\!\left/ \!\raisebox{-1ex}{$216$}\right.}=\frac{1}{6}$

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In Anand district, Institute of Technology, Nirma University and Sardar Patel University (SPU) are considered among the best options for B.Tech. Also, Birla Vishvakarma Mahavidyalaya (BVM) is a top choice and well-known for its engineering programs with good placements and faculty.

7. The sample space of given experiment is

$S=\{HH,HT,TH,TT\}$

$\mathrm{(i)\; E}:\{HT,TH\}$

$\Rightarrow F:\{HT,TH\}$

$\Rightarrow E\cap F=\{HT,TH\}$

$P(F)=\frac{2}{4}=\frac{1}{2}$ $;P(E\cap F)=\frac{2}{4}=\frac{1}{2}$

$P(E/F)=\frac{P(E\cap F)}{P(F)}=\frac{\frac{1}{2}}{\frac{1}{2}}=1$

$\mathrm{(ii)\; E}:\left\{HH\right\}$

$F:\left\{TT\right\}$

$E\cap F:\left\{\right\}$

So, $,P(F)=\frac{1}{4}$ $,P(E\cap F)=\frac{0}{4}=0$

$P(E/F)=\frac{P(E\cap F)}{P(F)}=\frac{0}{\frac{1}{4}}=0$

IHM Bangalore accepts scores obtained in the NHMCT-JEE exam for admission to its BSc in HHA course. Thus, students aspiring to take admission in the course must appear for the exam. The allotment of IHM Bangalore BSc seats is facilitated by NTA via the NHMCT counselling process. Only registered candidates are included in the rank list and can participate in the seat allotment process.

Yes, there are many government Communication Design colleges in Delhi NCR. Some of them are mentioned below along with their tuition fees:

Disclaimer: This information is sourced from the official website/ sanctioning body and may vary.

6. The sample space has 8 outcome,

$S=\{HHH,HHT,HTH,HTT,THH,THT,TTH,TTT\}$

$\mathrm{(i)\; E}=\{HHH,HTH,THH,TTH\}$

$F=\{HHH,HHT\}$

and $E\cap F=\left\{HHH\right\}$

$P(F)=\frac{2}{8}$ $;P(E\cap F)=\frac{1}{8}$

$P(E/F)=\frac{P(E\cap F)}{P(F)}=\frac{\frac{1}{8}}{\frac{2}{8}}=\frac{1}{2}$

$\mathrm{(ii)\; E}=\{HHH,HHT,HTH,THH\}$

$F=\{HHT,HTH,HTT,THH,THT,TTH,TTT\}$

$E\cap F=\{HHH,HTH,THH\}$

$P(F)=\frac{7}{8}$ $;P(E\cap F)=\frac{3}{8}$

$P(E/F)=\frac{P(E\cap F)}{P(F)}=\frac{\frac{3}{8}}{\frac{7}{8}}=\frac{3}{7}$

$\mathrm{(iii)\; E}:\{HHH,HHT,HTT,HTH,THH,THT,TTH\}$

$F:\{HHT,HTH,HTT,THH,THT,TTT,TTH\}$

$E\cap F:\{HHT,HTH,HTT,THH,THT,TTH\}$

$P(F)=\frac{7}{8}$ $;P(E\cap F)=\frac{6}{8}$

$P(E/F)=\frac{P(E\cap F)}{P(F)}=\frac{\frac{5}{8}}{\frac{6}{8}}=\frac{6}{7}$

No, CAT is not compulsory to get admissions to MBA courses at COER University. Because the university also considers the relevant scores from theMAT/XMAT/COERU Entrance test, followed by GD and PI. Therefore, candidates with valid scores in CAT can get admissions to MBA at COER University. However, it is not compulsory that candidates can only get admission through CAT. 

5. Given,

$P(A)=\frac{6}{11}$

$P(B)=\frac{5}{11}$

$P(A\cup B)=\frac{7}{11}$

$\mathrm{(i)\; P}(A\cap B)$

$P(A\cup B)=P(A)+P(B)-P(A\cap B)$

$\Rightarrow \frac{7}{11}=\frac{6}{11}+\frac{5}{11}-P(A\cap B)$

$\Rightarrow \frac{7}{11}-\frac{11}{11}=-P(A\cap B)$

$\Rightarrow -\frac{4}{11}=-P(A\cap B)$

$\Rightarrow P(A\cap B)=\frac{4}{11}$

$\mathrm{(ii)\; P}(A/B)$

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

$=\frac{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$11$}\right.}{\raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$11$}\right.}=\frac{4}{5}$

$\mathrm{(iii)\; P}(B/A)$

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

$=\frac{\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$11$}\right.}{\raisebox{1ex}{$6$}\!\left/ \!\raisebox{-1ex}{$11$}\right.}=\frac{4}{6}=\frac{2}{3}$

Yes, there are many Communication Design colleges in Delhi NCR. Some of them are mentioned below along with their tuition fees:

Disclaimer: This information is sourced from the official website/ sanctioning body and may vary.

Aditya Birla Education Academy admissions for all courses are currently ongoing. Interested candidates can apply in online mode through the official website. The online application process is explained in the following steps:

Step 1: Visit the official website of Aditya Birla Education Academy.

Step 2: Click on the register button.

Step 3: Fill out the form and submit it.