There are 12 (4 x 3) possible outcomes as listed below
(1,2), (1,4), (1,6), (3, 2), (3,4), (3, 6), (5, 2), (5,4), (5,6), (7,2). (7, 4) (7, 6) (ii) (a) only (1, 6), (3,.4), (5, 2) given sum as 7
\(\therefore\)Required probability = \(\frac {2} {13}=\frac {1} {4}\) 
(b) There is no even sum
\(\therefore\)Required probability= \(\frac {0} {12}=0\)
(c) All the sums are odd
\(\therefore\)Required probability =\(\frac {12} {12} =1\)
(d) only (3, 6), (5, 4), (5, 6), (7, 2). (7, 4) and (7, 6) gives sum as more than 7
\(\therefore\)Required probability = \(\frac {6} {12}\)
=\(\frac {1} {2}\)

Let G₁, G₂, G₃ and B₁, B₂ be three girls and two boys respectively.
Since two children are selected at random, therefore the following are the possible groups:
B₁B₂, B₁G₁, B₁G₂, B₁G₃, B₂G₁, B₂G₂, B₂G₃, G₁G₂, G₁G₃, G₂G₃
Total number of cases = 10
Now, at least one boy is selected in the following ways:
B₁B₂, B₁G₁, B₁G₂, B₁G₃, B₂G₁, B₂G₂, B₂G₃
Total number of favourable cases = 7  
\(\therefore\) Required probability = \(​​\frac {7} {10}\) 

Since the two dice are thrown simultarrously Total number Of outconrs = 6 x 6 = 36
Number of outconrs for getting same numbers on both dice = 6
\(\Rightarrow\)P(getting same number) = \(​​\frac { 6 }{ 36 } = \frac { 1 }{ 6 } \)
Now, P(getting same numbers) + P(getting different numbers) = 1
\(\Rightarrow \frac{ 1 } {6}\) + P(getting different numbers)= 1
\(\Rightarrow\)P(gerting different numbers) = 1-\(\frac {1} {6} = \frac {5} {6}\)

Here, total number of marbles = 12
Number of white marbles = y
(i) P(a white marble) = \(\frac { y }{ 12 } \)
(ii) Now, 6 more white marbles are added in the bag.
\(\therefore\)Total number Of marbles = 12 + 6 = 18
Number of white marbles = y + 6 .
\(\therefore\) P(a white marble) = \(\frac { y+6 }{ 18 } \)
According to the statement of the question, we have
\(​​\frac { y+6 }{ 18 } =2(\frac { y }{ 12 } )\)
\(\Rightarrow ​​\frac { y+6 }{ 3 } =y\)
\(\Rightarrow \) 3y=y+6
\(\Rightarrow \) 2y=6
\(\Rightarrow \) y=3