Aptitude - Probability

Answer: Option E

Explanation:

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

Answer: Option D

Explanation:

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

Answer: Option A

Explanation:

P (getting a prize) =	10	=	10	=	2	.
	(10 + 25)		35		7

Answer: Option C

Explanation:

Let S be the sample space.

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

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

n(E) = 4C2 =	(4 x 3)	= 6.
	(2 x 1)

P(E) =	n(E)	=	6	=	1	.
	n(S)		1326		221

Answer: Option D

Explanation:

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