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

 

Here S = {1,2,3,4,5,6}

Let E be the event of getting the multiple of 3

Then, E = {3,6}

P(E) = n(E)/n(S) = 2/6 = 1/3 

Here, S={HH,HT,TH,TT}
Let E be the event of getting one head
E={TT,HT,TH}
P(E )= n(E)/n(S) =3/4

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

Total number of elementary events = 20C2 ways =190

There are 7 white balls out of which one white can be drawn in 7C1 ways and one ball from remaining 13 balls can be drawn in 13C1 ways.

So,favourable events = 7C1*13C1

Therfore,required probability = (7C1*13C1)/20C2=91/90

Total number of elementary events = 52

There are 26 red cards,out of which one red card can be drawn in 26C1 ways =26.

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

Total number of elementary events = 10C5

Number of ways of selecting no defective bulbs i.e., 5 non-defective bulbs out of 7 is 7C5.

So,required probability =7C5/ 10C5 = 1/12.

Total number of elementary events = 10C5

Number of ways of selecting exactly one defective bulb out of 3 and 4 non-defective out of 7 is 3C1*7C4

So,required probability =3C1*7C4/10C5 = 5/12.

Total number of elementary events = 50C5
Given,third ticket =30

=> first and second should come from tickets numbered 1 to 29 = 29C2 ways and remaining two in 20C2 ways.

Therfore,favourable number of events = 29C2*20C2

Hence,required probability = 29C2*20C2/50C5 =551 / 15134