In a family with 2 children there are four possibilities:
1) the first child is a boy and the second child is a boy (bb)
2) the first child is a boy and the second child is a girl (bg)
3) the first child is a girl and the second child is a boy (gb)
4) the first child is a girl and the second child is a girl (gg)

But already given that one child is boy. So we have three possibilities of (bb)(bg)(gb).
n(E)= both are boys=BB=1
n(S)= 3
Required probability P = n(E)/n(S) = 1/3.

The number of exhaustive outcomes is 36.
Let E be the event of getting an even number on one die and an odd number on the other. Let the event of getting either both even or both odd then = 18/36 = 1/2
P(E) = 1 - 1/2 = 1/2.

Number of ways of (selecting at least two couples among five people selected) = (?C? x ?C?)

As remaining person can be any one among three couples left.

Required probability = (?C? x ?C?)/¹?C?
= (10 x 6)/252 = 5/21

In a simultaneous throw of two dice, n(S) = 6 x 6 = 36

Let E = event of getting a doublet = { (1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}

  P ( E ) = n ( E ) n ( S ) = 6 36 = 1 6

Total number of possible ways = 

9 C 5 = 9 x 8 x 7 x 6 x 5 5 x 4 x 3 x 2 x 1 = 126

Number of favorable cases = 

$$\begin{gathered} 4 C 2 x 5 C 3 + 4 C 3 x 5 C 2 + 4 C 4 x 5 C 1 \\ 4 x 3 2 x 1 x 5 x 4 x 3 3 x 2 x 1 + 4 x 3 x 2 3 x 2 x 1 x 5 x 4 2 x 1 + 1 x 5 \\ = 6 x 10 + 4 x 10 + 5 \\ = 60 + 40 + 5 \\ = 105 \end{gathered}$$

Therefore, required probability = 105/126 = 5/6

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

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

E = {(6, 3), (5, 4), (4, 5), (3, 6) }

=> n(E) = 4

Therefore, P(E) = 4/36 = 1/9

Select a number which ocurs on two dice out of six numbers (1, 2, 3, 4, 5, 6). This can be done in  C 1 6, ways.

Now select two distinct number out of remaining 5 numbers which can be done in  C 2 5 ways. Thus these 4 numbers can be arranged in 4!/2! ways.

So, the number of ways in which two dice show the same face and the remaining two show different faces is 

  C 1 6 × C 2 5 × 4 ! 2 ! = 720

 =>  n(E) = 720

  ? P E = 720 6 4 = 5 9

n(S) =  C 2 20 = 190 

n(E) =  C 2 15 = 105 

Therefore, P(E) = 105/190 = 21/38

 Vowels are A I A I O, 

 C    A   S    T   I    G   A    T   I    O   N

(O) (E) (O) (E) (O) (E) (O) (E) (O) (E) (O)

So there are 5 even places in which five vowels can be arranged and in rest of 6 places 6 constants can be arranged as follows :

n ( E ) = 5 P 5 2 ! * 2 ! ( A , I a r e 2 t i m e s ) * 6 P 6 2 ! ( T i s 2 t i m e s ) = 21600

n ( S ) = 11 ! 2 ! * 2 ! * 2 ! ( A , I a r e 2 t i m e s ) = 4989600

21600 4989600 = 0 . 043

Here, s={H,T} and E={H}
P(E) = n(E)/n(S) = 1/2