Total number of elementary events = 20 C 2 ways =190

There are 7 white balls out of which one white can be drawn in 7 C 1 ways and one ball from remaining 13 balls can be drawn in 13 C 1 ways.

So,favourable events =  7 C 1 * 13 C 1

Therfore,required probability = ( 7 C 1 * 13 C 1)/ 20 C 2=91/90

Probability of getting 6 at the top once =1/6

Probability of getting 6 at the top three times =1/6 x 1/6 x 1/6 =216

�Probability of no getting 6 at he top any time = 1 - 1/216 = 215/216

Let S be the sample space and E be the event of selecting 1 girl and 2 boys.

Then, n(S) = Number ways of selecting 3 students out of 25

                =  25 C 3  = 2300.

         n(E)= 10 C 1 * 15 C 2 = 1050. 

P(E) = n(E)/n(s) = 1050/2300 = 21/46

Total numbers in die=6

P(multiple of 2)=1/2

P(multiple of 5)=1/6

P(multiple of 2 or 5)=2/3

P(heart cards)=13/52

P(diamond cards)=13/52

P(heart or diamond)=13/52+13/52=1/2

P(red cards)=26/52

P(black cards)=26/52

P(red or black cards)=26/52+26/52=1

P(red ball) = 5/12

P(white ball) = 4/12

P(red or white ball) = 5/12 + 4/12 = 3/4 = 0.75

Let A be the event of first person hitting the target,

P ( A ) = 3 3 + 5 = 3 8 ( o d d i n f a v o u r )

Let B be the event of Second person hitting a target.

P ( B ) = 2 3 + 2 = 2 5 ( o d d a g a i n s t )

Since both events are independent and both will hit the target so,

P ( A ? B ) = P ( A ) P ( B ) = 3 8 � 2 5 = 3 20

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

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

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

P(getting prize) = 10/ (10 + 25) =2/7