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!

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!

Dear Student,

N(Sample space )=(1,2,3,4,5,6)

Probability of getting prime number on exactly two dice

N(A)=(1,1),(1,2),(1,3),(1,5),(2,1),(2,2),(2,3),(2,5),(3,1),(3,2),(3,3),(5,1),(5,2),(5,3),(5,5)

N(probability)=N(A)/N(s)

Since the question is about two dice

So N(s)=36

So answer will be 15/36

Hello Aspirant,

This is based on multiplication rule of probability.When we calculate probabilities of two independent events occurring at the same time, we need to simply multiply the two probabilities together. There are 9*3=27 ways of something happening or ways of outcome.

If, the first event happening affects the probability of the second event. It is called dependent events.In other cases, if the first event happening does not affect the probability of the second one. Then they are independent events.


You are throwing three dice, each of which has 6 possible outcomes.

The total number of outcomes is therefore 63= 6 x 6 x 6 =216

Each dice hasthreefavourable outcomes, 1, 2, or 3.

For the first two dice, you need to throw either 1, 2, or 3 for both dice. The favourable outcomes are:

1-11-21-32-12-22-33-13-23-3

In other words there areninefavourable outcomes with two dice. Now each one of these has three possible favourable outcomes from the third dice (ie. the third dice could be 1, 2, or 3).

So the number of favourable outcomes is 9 x 3 =27.
Hope this relates to the context of your question.

All the best