A sequence a, ax, ax2, ......, axn, has odd number of terms. Then the median is

 

Out of 9 persons,4 can be choosen in 9C4 ways =126.

Favourable events for given condition = 2C2*7C2= 21.

So,required probability = 21/126 =1/6.

3 letters can be choosen out of 9 letters in 9C3 ways.

More than one vowels ( 2 vowels + 1 consonant  or  3 vowels ) can be choosen in (4C2*5C1)+4C3 ways

Hence,required probability = 4C2*5C1+4C39C3 = 17/42

4 persons can be selected from 9 in 9C4 ways =126

Fvaourable events =4C2*5C2 =60

So,required probability = 60/126 = 10/21

Total number of persons = 9

Out of 9 persons 4 persons can be selected in 9C4 ways =126

1 man,1 woman and 2 children can be selected in 3C1*2C1*4C1 ways =36

So,required probability = 36/126 =2/7

Total number of events=52

Number of aces =4

So,required probability = 4/52 =1/13

Total number of events=6 x 6 x 6=216
Let A be the event of getting a total of atleast 6 and B denoted event of getting a total of less than 6 i.e.,3,4,5.

So,B={(1,1,1),(1,1,2),(1,2,1),(2,1,1),(1,1,3),(1,3,1),(3,1,1),(1,2,2),(2,1,2),(2,2,1)}

Favourable number of cases=10

Therefore,P(B)=10/216

=> P(A) = 1 - P(B)

             = 1 - (10/216) = 103/108

In a leap year,there are 366 days=52 weeks and 2 days

Remaining favourable 2 days can be sunday and monday or saturday and sunday

Exhaustive number of cases =7

Favourable number of cases =2

So,required probability=2/7

Here,n = 4(children)

P(girl)= 0.5

P(of atleast one girl)= 1 - P(no girls)

                             = 1 - 0.0625 = 0.9375