Two dice are tossed. The probability that the total score is a prime number is:

Here, S = {1, 2, 3, 4, ...., 19, 20}.

Let E = event of getting a multiple of 3 or 5 = {3, 6 , 9, 12, 15, 18, 5, 10, 20}.

∴ P(E) =	n(E)	=	9	.
	n(S)		20

Let S be the sample space.

Then, n(S)	= number of ways of drawing 3 balls out of 15
	= 15C3
	= (15 x 14 x 13) (3 x 2 x 1)
	= 455.

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

∴ n(E) = 5C3 = 5C2 =	(5 x 4)	= 10.
	(2 x 1)

∴ P(E) =	n(E)	=	10	=	2	.
	n(S)		455		91

Distance covered by B in 9 sec. =	❨	100	x 9	❩m	= 20 m.
		45

∴ A beats B by 20 metres.

A : B = 100 : 80.

A : C = 100 : 72.

∴

B

=

❨

B

x

A

❩

=

❨

80

x

100

❩

=

10

=

100

= 100 : 90.

C

A

C

100

72

9

90

∴ B can give C 10 points.

Ratio of the speeds of A and B =	5	: 1 = 5 : 3.
	3

Thus, in race of 5 m, A gains 2 m over B.

2 m are gained by A in a race of 5 m.

80 m will be gained by A in race of	❨	5	x 80	❩m	= 200 m.
		2

∴ Winning post is 200 m away from the starting point.

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

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

Let S be the sample space.

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

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 x 13C1)
	= (13 x 13)
	= 169.

∴ P(E) =	n(E)	=	169	=	13	.
	n(S)		1326		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)	=	2	=	1	.
	n(S)		52		26

In two throws of a dice, n(S) = (6 x 6) = 36.

Let E = event of getting a sum ={(3, 6), (4, 5), (5, 4), (6, 3)}.

∴ P(E) =	n(E)	=	4	=	1	.
	n(S)		36		9

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

Then, E

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

∴ n(E) = 27.

∴ P(E) =	n(E)	=	27	=	3	.
	n(S)		36		4