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Two dice are tossed. The probability that the total score is a prime number is:

A.

1/6

B.

1/2

C.

5/12

D.

3/4

Answer with explanation

Answer: Option CExplanation

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

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fifteen persons are sitting around a circular table facing the centre. What is the probability that three particular persons sit together ?

A.

3/91

B.

2/73

C.

1/91

D.

3/73

Answer with explanation

Answer: Option AExplanation

In a circle of n different persons, the total number of arrangements possible = (n – 1)!

Total number of arrangements = n(S) = (15 – 1)! = 14 !

Taking three persons as a unit, total persons = 13 (in 4 units)

Therefore no. of ways for these 13 persons to around the circular table = (13 – 1)! = 12!

In any given unit, 3 particular person can sit in 3!. Hence total number of ways that any three person can sit =

n(E) = 12! X 3!

Therefore P (E) = probability of three persons sitting together = n(E) / n(S) = 12! X 3 ! / 14!

= 12!x3x2 / 14x13x12! = 6/14×13 = 3/91

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The probabilities that a student will receive an A, B, C or D grade are 0.4, 0.3 , 0.2 and 0.1 respectively. Find the probability that a student will receive Atleast B grade.

A.

0.21

B.

0.3

C.

0.7

D.

None of these

Answer with explanation

Answer: Option CExplanation

P(atleast B) = P( B or A) = P(B) + P(A) = (0.3) + (0.4) = 0.7

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A brother and a sister appear for an interview against two vacant posts in an office. The probability of the brother’s selection is 1/5 and that of the sister’s selection is 1/3. What is the probability that only one of them is selected?

A.

1/5

B.

3/4

C.

2/5

D.

3/5

Answer with explanation

Answer: Option CExplanation

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If x is chosen at random from the set {1,2,3,4} and y is to be chosen at random from the set {5,6,7}, what is the probability that xy will be even?

A.

1/2

B.

2/3

C.

3/4

D.

4/3

Answer with explanation

Answer: Option DExplanation

S ={(1,5),(1,6),(1,7),(2,5),(2,6),(2,7),(3,5),(3,6),(3,7),(4,5),(4,6),(4,7)}

Total element n(S)=12

xy will be even when even x or y or both will be even.

Events of x, y being even is E.

E ={(1,6),(2,5),(2,6),(2,7),(3,6),(4,5),(4,6),(4,7)}

n(E) = 8

P(E)=n(E)n(S)=812 = 4/3

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An unbiased cubic die is thrown.What is the probabiltiy of getting a multiple of 3 or 4?

A.

1/12

B.

1/9

C.

3/4

D.

1/2

Answer with explanation

Answer: Option DExplanation

Total numbers in a die=6

P(mutliple of 3) = 2/6 = 1/3

P(multiple of 4) = 1/6

P(multiple of 3 or 4) = 1/3 + 1/6 = 1/2

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What is the probability that a leap year has 53 Saturdays and 52 Sundays ?

A.

1/7

B.

2/7

C.

1/2

D.

3/2

Answer with explanation

Answer: Option AExplanation

A leap year has 52 weeks and two days

Total number of cases = 7

Number of favourable cases = 1

i.e., {Friday, Saturday}

Required Probability = 1/7

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8 couples (husband and wife) attend a dance show “Nach Baliye’ in a popular TV channel ; A lucky draw in which 4 persons picked up for a prize is held, then the probability that there is atleast one couple will be selected is :

A.

8/39

B.

15/39

C.

12/13

D.

None of these

Answer with explanation

Answer: Option BExplanation

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A number X is chosen at random from the numbers -3, -2, -1, 0, 1, 2, 3. What is the probability that |X|<2

A.

3/7

B.

3/4

C.

4/5

D.

5/7

Answer with explanation

Answer: Option AExplanation

X can take 7 values.

To get |X|+2) take X={−1,0,1}

=> P(|X|<2) = Favourable CasesTotal Cases = 3/7

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I forgot the last digit of a 7-digit telephone number. If 1 randomly dial the final 3 digits after correctly dialing the first four, then what is the chance of dialing the correct number?

A.

1/999

B.

1/1001

C.

1/1000

D.

4/1000

Answer with explanation

Answer: Option CExplanation

It is given that last three digits are randomly dialled. then each of the digit can be selected out of 10 digits in 10 ways.

Hence required probability =(110)31103 = 1/1000

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There are four hotels in a town. If 3 men check into the hotels in a day then what is the probability that each checks into a different hotel?

A.

1/2

B.

3/4

C.

4/7

D.

3/8

Answer with explanation

Answer: Option DExplanation

Total cases of checking in the hotels = 4 x 4 x 4 = 64 ways.

Cases when 3 men are checking in different hotels = 4×3×2 = 24 ways.

Required probability =24/64 = 3/8

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