One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card (Jack, Queen and King only)?

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)	=	15	=	5	.
	n(S)		36		12

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.

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

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
	= 25C3 `
	= (25 x 24 x 23) (3 x 2 x 1)
	= 2300.

n(E)	= (10C1 x 15C2)
	= [ 10 x (15 x 14) ] (2 x 1)
	= 1050.

∴ P(E) =	n(E)	=	1050	=	21	.
	n(S)		2300		46