how many pairs of letters there in calendar
Hello candidate,
If we are talking about making pairs in the word "calendar", the total no of letters in the word are 8 and for forming a pair we need 2. So it can be done in 8c2 ways that is 28.
Thankyou!
Answer laterTwo numbers are selected at random from 123.....100 and are multiplied then the probability correct to two places of decimals that the product thus obtained is divisible by 3 is ????
Greetings of the day dear aspirant
Total cases = 100 C 2
Favorable cases = 33 C 1 . 67 C 1 + 33 C 2
(We know that from 1 to 100, 33 numbers are exactly divisible by 3, such that if any of these numbers are multiplied by remaining 67 numbers, their product should be divisible by 3, or any two numbers from 33 numbers are multiplied, their product should be divisible by 3)
Hence, the required probability = ( 33 C 1 . 67 C 1 + 33 C 2 )/ 100 C 2
= [(33. 67) +528]/ 4950
=(2211+528)/4950
= 2739/4950
= 0.55.
Hope this helps!!
All the best for your future
A number is chosen randomly from numbers 1 to 60.The probability that the chosen number is a multiple of 2 or 5 is.?
Hey there,
Hope you are doing well!
Let us suppose A:Number is multiple of 2.
Let us suppose B:number is multiple of 5.
Then, n(A)=30
n(B)=12
n(A intersection B)=6
=> P(A or B)=P(A)+P(B)-P(A and B)
=> P(A or B)= 30/60 + 12/60 -6/60
=> P(A or B)=36/60
=> P(A or B)= 3/5 ans.
Hope it helps!
are mathematical reasoning, induction, probability, and statistics important for Jee mains ?
In JEE mains exam you have to solve one question from Mathematical Reasoning of 4 marks and keep in mind that mathematical reasoning is not a part of JEE advance.
From Statisticsand Probability there will be 2 questions.
You can afford to neglect the mathematical reasoning part if you have time constraints.
You can click on the link below to check entire weightage list of jee mains syllabus.
https://www.google.com/amp/s/engineering.careers360.com/articles/jee-main-syllabus-weightage/amp
I hope this helps.
given the following probability distribution x compute i)E(X) ii)E(2x + or- 3) iii)V(X) iv)V(2x + or - 3) x. -3 -2. -1 0 1 2 3 f(x). 0.05 0.10 0.30. 0. 0.30 0.15 0.10
Hello
E(x) is simply given by summing x*f(x) for all x in case of discrete distribution like this one.This gives
E(x)= (-3)*(.05)+(-2)(0.10)+(-1)(0.30)+0+1(0.30)+2(0.15)+3(0.10)= 0.25
Since E(x) is a linear transformation, we can write
E(2x+3)= 2E(x)+ 3 = 3.5
and E(2x-3)= -2.5
how many questions can we expect from probability in viteee
Hello,
you can expect 2-3 questions. probability is also related to permutation and combination. so it can be a mixed question too. It is hard to get the accurate number. paper is set by the board and no one knows what will come until paper is distributed in the exam hall.
Keep on preparing. Solve as many questions you can. Appear mock tests. solve previous year papers. Check your performance and work to improve weak sections.
Hope it helps!
