**Directions (1-3): An urn contains some balls colored white, blue and green. The probability of choosing a white ball is 4/15 and the probability of choosing a green ball is 2/5. There are 10 blue balls.**

- What is the probability of choosing one blue ball?

A) 2/7

B) 1/4

C) 1/3

D) 2/5

E) 3/7

View Answer

** Option C**

** Solution: **

Probability of choosing one blue ball = 1 – (4/15 + 2/5) = 1/3

- What is the total number of balls in the urn?

A) 45

B) 34

C) 40

D) 30

E) 42

View Answer

**Option D**

** Solution:**

Probability of choosing one blue ball is 1/3

And total blue balls are 10. So with 10/30 we get probability as 1/3

So total balls must be 30

- If the balls are numbered 1, 2, …. up to number of balls in the urn, what is the probability of choosing a ball containing a multiple of 2 or 3?

A) 3/4

B) 4/5

C) 1/4

D) 1/3

E) 2/3

View Answer

**Option E**

** Solution:**

There are 30 balls in the urn.

Multiples of 2 up to 30 = 30/2 = 15

Multiples of 3 up to 30 = 30/3 = 10 (take only whole number before the decimal part)

Multiples of 6 (2×3) up to 30 = 30/6 = 5

So total such numbers are = 15 + 10 – 5 = 20

So required probability = 20/30 = 2/3

- There are 2 brothers A and B. Probability that A will pass in exam is 3/5 and that B will pass in exam is 5/8. What will be the probability that only one will pass in the exam?

A) 12/43

B) 19/40

C) 14/33

D) 21/40

E) 9/20

View Answer

** Option B**

** Solution: **

Only one will pass means the other will fail

Probability that A will pass in exam is 3/5. So Probability that A will fail in exam is 1 – 3/5 = 2/5

Probability that B will pass in exam is 5/8. So Probability that B will fail in exam is 1 – 5/8 = 3/8

So required probability = P(A will pass)*P(B will fail) + P(B will pass)*P(A will fail)

So probability that only one will pass in the exam = 3/5 * 3/8 + 5/8 * 2/5 = 19/40

- If three dices are thrown simultaneously, what is the probability of having a same number on all dices?

A) 1/36

B) 5/36

C) 23/216

D) 1/108

E) 17/216

View Answer

**Option A**

** Solution:**

Total events will be 6*6*6 = 216

Favorable events for having same number is {1,1,1}, {2,2,2}, {3,3,3}, {4,4,4}, {5,5,5}, {6,6,6} – so 6 events

Probability of same number on all dices is 6/216 = 1/36

- There are 150 tickets in a box numbered 1 to 150. What is the probability of choosing a ticket which has a number a multiple of 3 or 7?

A) 52/125

B) 53/150

C) 17/50

D) 37/150

E) 32/75

View Answer

**Option E**

** Solution:**

Multiples of 3 up to 150 = 150/3 = 50

Multiples of 7 up to 150 = 150/7 = 21 (take only whole number before the decimal part)

Multiples of 21 (3×7) up to 150 = 150/21 = 7

So total such numbers are = 50 + 21 – 7 = 64

So required probability = 64/150 = 32/75

- There are 55 tickets in a box numbered 1 to 55. What is the probability of choosing a ticket which has a prime number on it?

A) 3/55

B) 5/58

C) 8/21

D) 16/55

E) 4/13

View Answer

**Option D**

** Solution:**

Prime numbers up to 55 is 16 numbers which are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 43, 41, 47, 53.

So probability = 16/55

- A bag contains 4 white and 5 blue balls. Another bag contains 5 white and 7 blue balls. What is the probability of choosing two balls such that one is white and the other is blue?

A) 61/110

B) 59/108

C) 45/134

D) 53/108

E) 57/110

View Answer

**Option D**

** Solution:**

Case 1: Ball from first bag is white, from another is blue

So probability = 4/9 * 7/12 = 28/108

Case 1: Ball from first bag is blue, from another is white

So probability = 5/9 * 5/12 = 25/108

Add the cases

So required probability = 28/108 + 25/108 = 53/108

- The odds against an event are 2 : 3 and the odds in favor of another independent event are 3 : 4. Find the probability that at least one of the two events will occur.

A) 11/35

B) 27/35

C) 13/35

D) 22/35

E) 18/35

View Answer

**Option B**

** Solution:**

Let 2 events A and B

Odds against A are 2 : 3

So probability of occurrence of A = 3/(2+3) = 3/5. And non-occurrence of A = 2/5

Odds in favor of B are 3 : 4

So probability of occurrence of B = 3/(3+4) = 3/7. And non-occurrence of B = 4/7

Probability that at least one occurs

Case 1: A occurs and B does not occur

So probability = 3/5 * 4/7 = 12/35

Case 2: B occurs and A does not occur

So probability = 3/7 * 2/5 = 6/35

Case 3: Both A and B occur

So probability = 3/5 * 3/7 = 9/35

So probability that at least 1 will occur = 12/35 + 6/35 + 9/35 = 27/35

- The odds against an event are 1 : 3 and the odds in favor of another independent event are 2 : 5. Find the probability that one of the event will occur.

A) 17/28

B) 5/14

C) 11/25

D) 9/14

E) 19/28

View Answer

**Option A**

** Solution:**

Let 2 events A and B

Odds against A are 1 : 3

So probability of occurrence of A = 3/(1+3) = 3/4. And non-occurrence of A = 1/4

Odds in favor of B are 2 : 5

So probability of occurrence of B = 2/(2+5) = 2/7. And non-occurrence of B = 5/7

Case 1: A occurs and B does not occur

So probability = 3/4 * 5/7 = 15/28

Case 2: B occurs and A does not occur

So probability = 2/7 * 1/4 = 2/28

So probability that one will occur = 15/28 + 2/28 = 17/28