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!

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

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!

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

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!