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

 

The total number of elementary events associated to the random experiments of throwing four dice simultaneously is:

= 6*6*6*6=64

n(S) = 64

Let X be the event that all dice show the same face. 

X = { (1,1,1,1,), (2,2,2,2), (3,3,3,3), (4,4,4,4), (5,5,5,5), (6,6,6,6)}

n(X) = 6

Hence required probability = n(X)n(S)=664 =1216

Let A be the event that X is selected and B is the event that Y is selected.

P(A) = 1/7,  P(B) = 2/9.

Let C be the event that both are selected.

P(C) = P(A) × P(B) as A and B are independent events: 

       = (1/7) × (2/9)  = 2/63

X can take 7 values.
To get |X|+2) take X={−1,0,1}

=> P(|X|<2) = Favourable CasesTotal Cases = 3/7

One person can select one house out of 3= 3C1 ways =3.

Hence, three persons can select one house out of 3 in 3 x 3 x 3 =9.

Therefore, probability that all thre apply for the same house is 1/9

We have n(s) =52C2 52 = 52*51/2*1= 1326. 

Let A = event of getting both black cards 

     B = event of getting both queens 

A∩B = event of getting queen of black cards 

n(A) = 52*512*1 = 26C2 = 325, n(B)= 26*252*1= 4*3/2*1= 6  and  n(A∩B) = 4C2 = 1 

P(A) = n(A)/n(S) = 325/1326;

P(B) = n(B)/n(S) = 6/1326 and 

P(A∩B) = n(A∩B)/n(S) = 1/1326 

P(A∪B) = P(A) + P(B) - P(A∩B) = (325+6-1) / 1326 = 330/1326 = 55/221

Clearly, n(S) = 6 x 6 = 36
Let E be the event that the sum of the numbers on the two faces is divided by 4 or 6.
Then,E = {(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5),(4,2),(4,4),(5,1),(5,3),(6,2),(6,6)}
n(E) = 14.
Hence, P(E) = n(E)/n(S) = 14/36 = 7/18

Let S be the sample space

Then n(S) = no of ways of drawing 2 balls out of (6+4) =10C2 10 =10*92*1 =45

Let E = event of getting both balls of same colour

Then,n(E) = no of ways (2 balls out of six) or (2 balls out of 4)

                =6C2+4C2 = 6*52*1+4*32*1= 15+6 = 21

Therefore, P(E) = n(E)/n(S) = 21/45 = 7/15

Here n(S) = (6 x 6) = 36

Let E = event of getting a total more than 7
        = {(2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6)}

Therefore,P(E) = n(E)/n(S) = 15/36 = 5/12.