Sasi Math (Revised for 1024)

Scroll:Probability >> probability >> saq (4236)

Written Instructions:

For each Multiple Choice Question (MCQ), four options are given. One of them is the correct answer. Make your choice (1,2,3 or 4). Write your answers in the brackets provided..

For each Short Answer Question(SAQ) and Long Answer Question(LAQ), write your answers in the blanks provided.

Leave your answers in the simplest form or correct to two decimal places.

A bag contains 13 white,7 black, 4 green and 2 red balls. One ball is drawn at random.Find the probability that the ball drawn is white or black or green.

Answer:_______________

A bag contains 10 white,8 black, 5 green and 2 red balls. One ball is drawn at random.Find the probability that the ball drawn is white or black or green.

Answer:_______________

A bag contains 13 white,9 black, 6 green and 2 red balls. One ball is drawn at random.Find the probability that the ball drawn is white or black or green.

Answer:_______________

A bag contains 15 white,8 black, 4 green and 3 red balls. One ball is drawn at random.Find the probability that the ball drawn is white or black or green.

Answer:_______________

A bag contains 14 white,9 black, 5 green and 3 red balls. One ball is drawn at random.Find the probability that the ball drawn is white or black or green.

Answer:_______________

A bag contains 15 white,9 black, 6 green and 2 red balls. One ball is drawn at random.Find the probability that the ball drawn is white or black or green.

Answer:_______________

A bag contains 12 white,7 black, 6 green and 3 red balls. One ball is drawn at random.Find the probability that the ball drawn is white or black or green.

Answer:_______________

A bag contains 14 white,7 black, 6 green and 2 red balls. One ball is drawn at random.Find the probability that the ball drawn is white or black or green.

Answer:_______________

A bag contains 10 white,9 black, 5 green and 2 red balls. One ball is drawn at random.Find the probability that the ball drawn is white or black or green.

Answer:_______________

10)

A bag contains 11 white,9 black, 5 green and 3 red balls. One ball is drawn at random.Find the probability that the ball drawn is white or black or green.

Answer:_______________

A bag contains 13 white,7 black, 4 green and 2 red balls. One ball is drawn at random.Find the probability that the ball drawn is white or black or green.

Answer: $\frac{12}{13}$

SOLUTION 1 :

Let S be the sample space.

n(S) = 26

Let W, B and G be the events of selecting a white, black and green ball respectively .

Probability of getting a white ball, P(W)

P(W)	=		n(W)	=	13
			n(S)		26

Probability of getting a black ball,P(B)

P(B)	=	n(B)	=	7
		n(S)		26

Probability of getting a green ball,P(G)

P(G)	=	n(G)	=	4
		n(S)		26

Probability of getting a white or black or green ball,

P(W ∪ B ∪ G) = P(W) + P(B) + P(G) W,B,G are mutually exclusive.

=	13	+	7	+	4	=	24
	26		26		26		26

ans = $\frac{12}{13}$

A bag contains 10 white,8 black, 5 green and 2 red balls. One ball is drawn at random.Find the probability that the ball drawn is white or black or green.

Answer: $\frac{23}{25}$

SOLUTION 1 :

Let S be the sample space.

n(S) = 25

Let W, B and G be the events of selecting a white, black and green ball respectively .

Probability of getting a white ball, P(W)

P(W)	=		n(W)	=	10
			n(S)		25

Probability of getting a black ball,P(B)

P(B)	=	n(B)	=	8
		n(S)		25

Probability of getting a green ball,P(G)

P(G)	=	n(G)	=	5
		n(S)		25

Probability of getting a white or black or green ball,

P(W ∪ B ∪ G) = P(W) + P(B) + P(G) W,B,G are mutually exclusive.

=	10	+	8	+	5	=	23
	25		25		25		25

ans = $\frac{23}{25}$

A bag contains 13 white,9 black, 6 green and 2 red balls. One ball is drawn at random.Find the probability that the ball drawn is white or black or green.

Answer: $\frac{14}{15}$

SOLUTION 1 :

Let S be the sample space.

n(S) = 30

Let W, B and G be the events of selecting a white, black and green ball respectively .

Probability of getting a white ball, P(W)

P(W)	=		n(W)	=	13
			n(S)		30

Probability of getting a black ball,P(B)

P(B)	=	n(B)	=	9
		n(S)		30

Probability of getting a green ball,P(G)

P(G)	=	n(G)	=	6
		n(S)		30

Probability of getting a white or black or green ball,

P(W ∪ B ∪ G) = P(W) + P(B) + P(G) W,B,G are mutually exclusive.

=	13	+	9	+	6	=	28
	30		30		30		30

ans = $\frac{14}{15}$

A bag contains 15 white,8 black, 4 green and 3 red balls. One ball is drawn at random.Find the probability that the ball drawn is white or black or green.

Answer: $\frac{9}{10}$

SOLUTION 1 :

Let S be the sample space.

n(S) = 30

Let W, B and G be the events of selecting a white, black and green ball respectively .

Probability of getting a white ball, P(W)

P(W)	=		n(W)	=	15
			n(S)		30

Probability of getting a black ball,P(B)

P(B)	=	n(B)	=	8
		n(S)		30

Probability of getting a green ball,P(G)

P(G)	=	n(G)	=	4
		n(S)		30

Probability of getting a white or black or green ball,

P(W ∪ B ∪ G) = P(W) + P(B) + P(G) W,B,G are mutually exclusive.

=	15	+	8	+	4	=	27
	30		30		30		30

ans = $\frac{9}{10}$

A bag contains 14 white,9 black, 5 green and 3 red balls. One ball is drawn at random.Find the probability that the ball drawn is white or black or green.

Answer: $\frac{28}{31}$

SOLUTION 1 :

Let S be the sample space.

n(S) = 31

Let W, B and G be the events of selecting a white, black and green ball respectively .

Probability of getting a white ball, P(W)

P(W)	=		n(W)	=	14
			n(S)		31

Probability of getting a black ball,P(B)

P(B)	=	n(B)	=	9
		n(S)		31

Probability of getting a green ball,P(G)

P(G)	=	n(G)	=	5
		n(S)		31

Probability of getting a white or black or green ball,

P(W ∪ B ∪ G) = P(W) + P(B) + P(G) W,B,G are mutually exclusive.

=	14	+	9	+	5	=	28
	31		31		31		31

ans = $\frac{28}{31}$

A bag contains 15 white,9 black, 6 green and 2 red balls. One ball is drawn at random.Find the probability that the ball drawn is white or black or green.

Answer: $\frac{15}{16}$

SOLUTION 1 :

Let S be the sample space.

n(S) = 32

Let W, B and G be the events of selecting a white, black and green ball respectively .

Probability of getting a white ball, P(W)

P(W)	=		n(W)	=	15
			n(S)		32

Probability of getting a black ball,P(B)

P(B)	=	n(B)	=	9
		n(S)		32

Probability of getting a green ball,P(G)

P(G)	=	n(G)	=	6
		n(S)		32

Probability of getting a white or black or green ball,

P(W ∪ B ∪ G) = P(W) + P(B) + P(G) W,B,G are mutually exclusive.

=	15	+	9	+	6	=	30
	32		32		32		32

ans = $\frac{15}{16}$

A bag contains 12 white,7 black, 6 green and 3 red balls. One ball is drawn at random.Find the probability that the ball drawn is white or black or green.

Answer: $\frac{25}{28}$

SOLUTION 1 :

Let S be the sample space.

n(S) = 28

Let W, B and G be the events of selecting a white, black and green ball respectively .

Probability of getting a white ball, P(W)

P(W)	=		n(W)	=	12
			n(S)		28

Probability of getting a black ball,P(B)

P(B)	=	n(B)	=	7
		n(S)		28

Probability of getting a green ball,P(G)

P(G)	=	n(G)	=	6
		n(S)		28

Probability of getting a white or black or green ball,

P(W ∪ B ∪ G) = P(W) + P(B) + P(G) W,B,G are mutually exclusive.

=	12	+	7	+	6	=	25
	28		28		28		28

ans = $\frac{25}{28}$

A bag contains 14 white,7 black, 6 green and 2 red balls. One ball is drawn at random.Find the probability that the ball drawn is white or black or green.

Answer: $\frac{27}{29}$

SOLUTION 1 :

Let S be the sample space.

n(S) = 29

Let W, B and G be the events of selecting a white, black and green ball respectively .

Probability of getting a white ball, P(W)

P(W)	=		n(W)	=	14
			n(S)		29

Probability of getting a black ball,P(B)

P(B)	=	n(B)	=	7
		n(S)		29

Probability of getting a green ball,P(G)

P(G)	=	n(G)	=	6
		n(S)		29

Probability of getting a white or black or green ball,

P(W ∪ B ∪ G) = P(W) + P(B) + P(G) W,B,G are mutually exclusive.

=	14	+	7	+	6	=	27
	29		29		29		29

ans = $\frac{27}{29}$

A bag contains 10 white,9 black, 5 green and 2 red balls. One ball is drawn at random.Find the probability that the ball drawn is white or black or green.

Answer: $\frac{12}{13}$

SOLUTION 1 :

Let S be the sample space.

n(S) = 26

Let W, B and G be the events of selecting a white, black and green ball respectively .

Probability of getting a white ball, P(W)

P(W)	=		n(W)	=	10
			n(S)		26

Probability of getting a black ball,P(B)

P(B)	=	n(B)	=	9
		n(S)		26

Probability of getting a green ball,P(G)

P(G)	=	n(G)	=	5
		n(S)		26

Probability of getting a white or black or green ball,

P(W ∪ B ∪ G) = P(W) + P(B) + P(G) W,B,G are mutually exclusive.

=	10	+	9	+	5	=	24
	26		26		26		26

ans = $\frac{12}{13}$

10)

A bag contains 11 white,9 black, 5 green and 3 red balls. One ball is drawn at random.Find the probability that the ball drawn is white or black or green.

Answer: $\frac{25}{28}$

SOLUTION 1 :

Let S be the sample space.

n(S) = 28

Let W, B and G be the events of selecting a white, black and green ball respectively .

Probability of getting a white ball, P(W)

P(W)	=		n(W)	=	11
			n(S)		28

Probability of getting a black ball,P(B)

P(B)	=	n(B)	=	9
		n(S)		28

Probability of getting a green ball,P(G)

P(G)	=	n(G)	=	5
		n(S)		28

Probability of getting a white or black or green ball,

P(W ∪ B ∪ G) = P(W) + P(B) + P(G) W,B,G are mutually exclusive.

=	11	+	9	+	5	=	25
	28		28		28		28

ans = $\frac{25}{28}$