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

 

Let number of balls = (6 + 8) = 14.

Number of white balls = 8.

P (drawing a white ball) = 8 /14=4/7.

Clearly, there are 52 cards, out of which there are 12 face cards.

P (getting a face card) = 12/52=3/13.

Let S be the sample space.

Then, n(S) = number of ways of drawing 3 balls out of 15 = 15C3  =15*14*133*2*1= 455.  

Let E = event of getting all the 3 red balls.

 n(E) = 5C3 = 5*42*1 = 10.

 => P(E) = n(E)/n(S) = 10/455 = 2/91.

Let S be the sample space.

Then, n(S) = 52C2=(52 x 51)/(2 x 1) = 1326.

Let E = event of getting 1 spade and 1 heart.

n(E)= number of ways of choosing 1 spade out of 13 and 1 heart out of 13 = 13C1*13C1 = 169.

P(E) = n(E)/n(S) = 169/1326 = 13/102.

Here, n(S) = 52.

Let E = event of getting a queen of club or a king of heart.

Then, n(E) = 2.

P(E) =n(E)/n(S)=2/52=1/26.

Clearly, n(S) = (6 x 6) = 36.

Let E = Event that the sum is a prime number.

Then E= { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4,3),(5, 2), (5, 6), (6, 1), (6, 5) }

n(E) = 15.

P(E) = n(E)/n(S) = 15/36 = 5/12.

 Let S be the sample space

Then, n(S) = 52C2= 1326.

Let E = event of getting 2 kings out of 4.

n(E)  = 52*512*1= 6.

 4C2

=>4*32*1

P(E)=n(E)n(S)=61326=1221

Total number of outcomes possible, n(S) = 10 + 25 = 35

Total number of prizes, n(E) = 10

 P(E)=n(E)n(S)=1035=27