In two throws of a die, n(s)=(6 x 6)=36

let E= Event of geting a sum 9={(3,6),(4,5),(5,4),(6,3)}

P(E) = n(E)/n(S) = 4/36 = 1/9

These two events cannot be disjoint because P(K) + P(L) > 1.

P(A?B) = P(A) + P(B) - P(A?B).

An event is disjoint if P(A ? B) = 0. If K and L are disjoint P(K ? L) = 0.8 + 0.6 = 1.4

And Since probability cannot be greater than 1, these two mentioned events cannot be disjoint.

The two events mentioned are independent.

The first roll of the die is independent of the second roll. Therefore the probabilities can be directly multiplied.

P(getting first 4) = 1/6

P(no second 6) = 5/6

Therefore P(getting first 4 and no second 6) = 1/6 x 5/6 = 5/36

S = { 1, 2, 3, 4, 5, 6 } 

=> n(S) = 6

E = { 3, 6}

=> n(E) = 2

Therefore, P(E) = 2/6 =1/3

Total number of cases=36

Number of favourable cases(sum 5 or 6)=10

P(getting total of 5 or 6)=10/36=5/18

Number of queen cards = 4

Number of ace cards = 4

 P(either a queen or ace) = 4/52 + 4/52= 8/52 = 2/13

Here,n = 4(children)

P(girl)= 0.5

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

                             = 1 - 0.0625 = 0.9375

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

Fvaourable events = 4 C 2 * 5 C 2 =60

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

Total number of elementary events = 52

There are 26 red cards,out of which one red card can be drawn in 26 C 1 ways =26.

So,required probability = 26/52 = 1/2