Sir, i am Sourav S. Bag from Sundargarh,Odisha. My OJEE RANK (Gen- 942, Sc- 123) and My JEE MAINS RANK (Air- 6,71,289 ; Sc- 61,252). So at the time of counciling which Rank should i prefer to submit, to get a seat at Govt. College?
Hello aspirant,
Hope you are fine and healthy.
As your JEE Mains rank for SC category is 61252 which is not very promising.
I would suggest that you should go with OJEE SC rank 123 which is quite impressive and have a chance to get government institutions.
Here is a list of colleges where you can try for admission,
=> Institute of Management and Information Technology, Cuttack
=> Government College of Engineering, Kalahandi
=> Government College of Engineering, Keonjhar
=> Parala Maharaja Engineering College, Behampur
Here is a link, where you can get more information regarding engineering colleges which take admission through OJEE counselling,
Hope it helps you..
Best of luck..
Answer latera bag contain 5red 4black balls a second bag contain 3red and 6black balls one of the two bag is selected random without replacement both of which are found to be red find the probability that these two balls are drawn from second bag
Bag A=(4R+5B) balls
⇒ Bag B=(3R+6B) balls
⇒ Probability of red form bag A and black from bag B =
9
4
×
10
6
=
90
24
⇒ Probability of red from bag B and black from bag A =
10
3
×
9
5
=
90
15
⇒ P(balls of different colours)=P( red from bag A and black from bag B)+P( red from bag B and black from bag A)
=
90
26
+
90
15
=
90
41
All the best student,
Hope this helps you.
If a student came from village that so far to allotted exam centre. will they keep their things like bag, study material or mobile phone etc.. in the exam centre or nearby invigilators besides?
Dear Ashish,
In any entrance examination, the bag or study material is not allowed to carry with you in your examination hall. If you talk about NEET , they don't allow you to carry your phone too.
You can submit your phone at the gate or entrance of examination centre. Same goes with any study material or bag, you need to submit it at the main gate they will provide you with the storage information, but you can't take any belongings with you inside your examination hall.
All you can take is your admit card.
The invigilator is not allowed to take responsibility of your belongings of any sort. So, you can't keep it to him/her.
Moreover, all the information is mentioned in the notification of NEET exam or you will also get all the information that is what to carry, what not to in your admit card as well. Yo u can even visit their website-
https://neet.nta.nic.in/
Further if you want to know more about NEET and want to enchance your NEET preparation, you can visit-
https://learn.careers360.com/knockoutfull-neet-may/
Hope this helped.
All the Best!!
i have joined BAG last year 2020 june session at IGNOU but i have not able to exam form fill up yet when the exam from fill up date will be announced.please guide me
Hello,
I would like to inform you that the TEE ( Term End Examination) for various degree courses for July -2020 cycle of IGNOU has been done already and the results have also been declared.
So if you due to ignorance or due to any other reason you couldn't appear for the examination then you can appear for the same examination in December 2021, for which registration will commence in upcoming weeks.
Also it is advised that you keep on checking the official website of IGNOU for all the information as many times you wouldn't get the notice for all the events personally. So if you don't wish to miss out on something very important, you should time to time visit the official website.
http://ignou.ac.in/
Thank you.
In a bag there are some gold coins.In another bag there are 1/3rd extra gold coins as compared to first bag.If the difference in the number of gold coins in first and second bag is 5,then how many coins are there in the first bag?
Hello candidate,
In the given question its mentioned that there are two bags in which one bag has x number of gold coins, and the second back has one third more number of gold coins which makes the number is equal to 4x/3.
It's also given that the difference in the number of gold coins in both the back is equal to 5, so 4x/3- x/3 =5. So, the number of gold coins in the smaller lot of bags is equal to 15 and the second bag has a total of 20 gold coins.
Hope you found it helpful.
If you have any further queries, feel free to post it here!!
A bag contains 5 red and 3 black balls and another bag contains 2 red and 6 black balls. Two balls are drawn at random (without replacement) from one of the bags and both are found to be red. Find the probability that balls are drawn from the first bag.
hello Aspirant
I hope you are doing good. This question can be answered by applying base theorem
probabibility ofn choose 1st bag =1/2
probability of chossing 2nd bag=1/2
Probability of getting red ball from bag 1=5/8
Probability of getting red ball from bag2=1/4
Proability of having red ball form bag1
= (1/2*5/8)/((1/2*5/8)+(1/2*1/4))
Solve this now by your own that will be your required answer
Hope you found this answer helpful. Good luck for your future
What score is required to bag a seat in Jadavpur University of I belong from other state ?
Hello,
Jadavpur University is a government college in West Bengal. If you are from another state then you have to give WBJEE examination. And get chance to admission here all private college and Jadavpur University.
So if you want to get admission here you have to score in between 100 to 160 or more than that in WBJEE examination. Otherwise it's quite difficult to get admission in Jadavpur University. And your 12th marks should be 60% if you are general category. Otherwise if you are reserve category student then your marks 55% is require from 12th.
All the best.
I have completed 1 year of BAG in ignou, given TE exams also, I did re registration process & submitted the application & paid the 2nd year fees, but some problem with bank the payment didnt go through,how should I pay fees now for the 2nd year.in the re- registeration page it shows payment failed.
Hello,
If you already paid the fees but didn't paid it or any baking service problem then you should wait some times. Sometimes service problem occurs. So that moment your money will be back in your account or it's go the college account. You should talk to the bank directly and your college authority also. Then you can do anything after listening their suggestions.
All the best.
Four balls are drawn from a bag containing 5 black 6 white and 7 red balls. let X is number of white balls drawn. find probability mass function or probability distribution of X
Hello candidate,
In The question you had mentioned that there are 5 black 6 white and 7 red balls, from which 4 balls are needed to be drawn and X denotes the number of white balls drawn in several attempts.
There can be several cases,
Let X equal to 1, so the total ways of drawing are- 6C1 × 12C3.
Let X equal to 2, the total number of ways of drawing are- 6C2 × 12C2.
Let X equal to 3, the total number of ways of drawing are- 6C3 × 12C1.
Let X Equal to 4, the total number of ways of drawing are- 6C4.
Hope This answer was helpful!!
A Bag contain 5 black 6 white and 7 red balls four balls are drawn at random from it.If x denotes the number of white balls,then findE(x)
Total number of balls in bag = 18 out of which 6 are white balls and other balls count to a total of 12.
It is given that X is the number of the white balls drawn. And the white balls are drawn four in number.
Now, since nothing is given so, we can assume that ball are drawn without replacement.
So, expectation E(X) = summation (XP(X))
So, let P(X=0) = all drawn balls are non-white. So X = 0 so, E ( X = 0) = 0
Now, probability of drawing one white ball, i.e P(X=1) = 4 {(6/18) (12/17) (11/16) (10/15)} = 0.4314
Since we are not told any order so there can be four cases that a White ball is drawn in first attempt or in second attempt or in third attempt or in fourth attempt or you cam saya 4C1 cases.
Now, P( X=2 ) = 6 {(6/18) (5/17) (12/16) (11/15)} = 0.3235
P(X=3) = 4 {(6/18) (5/17) (4/16) (12/15)} = 0.0784
And, P(X=4) = {(6/18) (5/17) (4/16) (3/15)} = 0.0049
So, now E(X) = (1 P(X=1)) + (2 P(X=2)) + (3 P(X=3)) + (4 P(X=4))
So, after calculating the values we get E(X) = 1.3332
Note that expectation is not the probability, so it's value can be also greater than 1.
