- In how many ways can a number 6084 be written as a product of two different factors ?

A: 27

B: 26

C: 13

D: 14

Answer:C

Explanation: Ans is 13

lcm of 6084= 2^2*3^3*13^13

and it is a perfect sq no, i.e., 78, so we have a formula for finding factors as product of two distinct factors.

1/2[(2+1 )*(2+1)*(2+1)-1] = 13

2. The value of (1/5 12)^1/9 is:

A. ½

B.1/3

C.1/4

D.1/6

Answer:A

Explanation: (1/512)^1/9=(1/2^9)^1/9=(2^-9)^1/9=1/2

3. There are 4 boxes colored red, yellow, green and blue. If 2 boxes are selected, how many combinations are there for at least one green box or one red box to be selected?

A.5

B.7

C.8

D.10

Ans:A

At least 1 green or 1 red means

(1 green and 1 other) or (1 red and 1 other) or (1 green and 1 red)

Other means yellow and blue

So it becomes (1c1 * 2c1) + (1c1 * 2c1) + (2c2)

= 2+2+1 = 5

4. A quiz has one multiple choice question with answer choices a,b and c and two true/false question. what is the probability of answering all the three questions correctly by guessing

a) 1/5

b)1/4

c)1/3

d)1/12

Ans:D

Explanation: for multiple choice=1/3

for true/false =1/2

therefore probability=1/3*1/2*1/2=1/12

5. From a pack of 52cards four cards removed at random.in how many ways first and third place cards are drawnout such that both are black?

A)1592500

B)32000

C)152609

D)none

Ans:A

Explanation: There are 26 black cards to choose from on the first draw.

Now the second card can either be black or red.

If black, then there are 25 to choose from, and then for

the third card there will then be 24 black cards to choose from.

If red, then there are 26 reds to choose from, and then

for the third card there are then 25 black cards to

choose from. In either case, there are 49 cards left to from on the 4th draw.

So the number of ways we can do this is

26*(25*24 + 26*25)*49 = 1,592,500

6. unbiased coin tossed N times.probability of getting 4 tails equals probability of getting 7 tails. then probability of getting 2 tails

a)55/2048

b)3/96

c)1/1024

d)none

Answer:A

Explanation: Probability of getting 4 tails=nc4/2^n

Probability of getting 7 tails=nc7/2^n

nc4/2^n=nc7/2^n

nc4=nc7

n=11

Required probability=11c2/2^11=55/2048

- A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random from the bag. The probability that all of them are red, is:

A)1/13

B)2/13

C)1/26

D)1/52

Answer:C

Explanation: 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 |

- A box contains 20 electric bulbs, out of which 4 are defective. Two bulbs are chosen at random from this box. The probability that at least one of these is defective, is:

A. 4/19

B. 7/19

C. 12/19

D. 21/95

Answer:B

Explanation: P( None is defective)

= 16C2 / 20C2

= (16×15/2×1 ×2×1/20×19)

= 12/19.

P( at least one is defective)

= (1- 12/19)

= 7/19.

- In a class, 30% of the students offered English, 20% offered Hindi and 10% offered both. If a student is selected at random, what is the probability that he. has offered English or Hindi ?

A) ½

B) ¾

C) 4/5

D) 2/5

Answer:D

Explanation:

- Two letters are randomly chosen from the word TIME. Find the probability that the letters are T and M?

A) ¼

B) 1/6

C) 1/8

D) 4

Answer: B) 1/6

Explanation:

Required probability is given by P(E) = n(E)n(S) = 14C2 = 16

- If A and B are 2 independent events and P(A)=0.5 and P(B) = 0.4, find P(A/B):

A: 0.5

B: 0.4

C: 0.88

D: None of these

Correct Answer : A

Explanation: P(A/B) = P(A ∪ B)/P(B)

P(A ∪ B) = P(A)P(B) = 0:2/0.4 =0.5

- A five-digit number is formed by using digits 1,2,3,4 and 5 without repetition. What is the probability that the number is divisible by 4?

A)1/5

B)2/5

C)3/5

D)4/5

Answer:A

Solution:

A number divisible by 4 formed using the digits 1,2,3,4 and 5 has to have the last two digits 12 or24 or 32 or 52.

In each of these cases, the five digits number can be formed using the remaining 3 digits in 3! = 6 ways.

A number divisible by 4 can be formed in 6×4=24 ways.

Total number that can be formed using the digits 1,2,3,4 and 5 without repetition

=5!=120

Required probability = 24/120 = 1/5

13.A bag contains 5 oranges, 4 bananas and 3 apples. Rohit wants to eat a banana or an apple. He draws a fruit from the bag randomly. What is the probability that he will get a fruit of his choice?

A)1/12

B)7/12

C)5/12

D)none of these

Answer: Option B

Explanation :

Total fruits = 12 (5+4+3)

Chances of selecting banana or apple = (4+3) = 7

So, probability =7/12

- In a lottery, there are 10 prizes and 25 blanks. A lottery is drawn at random. What is the probability of getting a prize?

A)1/10

B)2/5

C)2/7

D)5/7

Answer:C

Explanation:

P (getting a prize) = | 10 | = | 10 | = | 2 | . |

(10 + 25) | 35 | 7 |

- There are two boxes A and B. Box A has three red and four blue balls. Box B has five red and two blue balls. Anya draws a ball from each bag randomly. What is the probability that both balls are red?

A: 4/7

B: 8/49

C: 7/8

D: 15/49

Answer:

Explanation: Box A has 3 red balls and box B has 5 red balls

(i.e.,) total red balls => n(red balls) = 8

Total number of balls in both the boxes => n(S) = 7+7 = 14

P(both balls are red) = 8/14 = 4/7

- Tickets numbered 1 to 20 are mixed up and then a ticket is drawn at random. What is the probability that the ticket drawn has a number which is a multiple of 3 or 5?

A)1/2

B)2/5

C)8/15

D)9/20

Answer:D

Explanation: 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 |

- Ravi has a bag full of 10 Nestle and 5 Cadbury chocolates. He draws two chocolates. What is the probability that he got at least one Nestle chocolate?

A)19/20

B)3/7

C)2/21

D)1/3

Answer:A

Explanation: Total number sample space two chocolates can be drawn in the way of n(S) = 15C2=105

At least one to be nestle. n(E) = 10C1 * 5C1 + 10C2 * 5C0 = 95

P(E) = 95/105 = 19/21

**18.**A bag contains 5 yellow and 4 brown pencils. If two pencils are drawn, what is the probability that the pencils are of the same colour?

A)2/9

B)4/9

C)7/2

D)7/8

Answer:B

Explanation: Probability of picking 2 brown pencils = 2/9

Probability of picking 2 pencils of same colour = 2/9 + 2/9= 4/9

- A single letter is drawn at random from the word, “ASPIRATION”, the probability that it is a vowel?

A. ½

B. 1/3

C. 3/5

D. 2/5

Answer:A

Explanation: A,E,I,O,U are vowels.

There are 10 alphabets in the word “ASPIRATION”.

Out of which A,I,A,I,O are vowels.

So there are 5 vowels out of 10 total alphabets.

Thus the probability of choosing a random letter and getting a vowel is 5/10.(because in the given word ,out of 10 alphabets , 5 are vowels).

5/10 can be simplified to 1/2.

20.The probability that a man can hit a target is 3/4. He tries 5 times. The probability that he will hit the target at least three times is:

A)291/364

B)371/464

C)471/502

D)459/512

Answer:D

Explanation: Required Probability

Given, n = 5 and r = 3

Then, Success P = 3/4

Failure, q = 1 – 3/4 = 1/4

Man hit the target thrice

= ^{5}c_{3} (3/4)^{3} (1/4)^{2} + ^{5}c_{4} (3/4)^{3} (1/4) + ^{5}c_{3}(3/4)^{5}

= (270/1024) + (405/1024) + (243/1024)

= 918/1024

= 459/512