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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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Three houses are available in a locality. Three persons apply for the houses. Each applies for one house without consulting others. The probability that all the three apply for the same house is :

A.

2/9

B.

1/9

C.

8/9

D.

7/9

Answer with explanation

Answer: Option BExplanation

One person can select one house out of 3= 3C13C1 ways =3.

Hence, three persons can select one house out of 3 in 3 x 3 x 3 =9.

Therefore, probability that all thre apply for the same house is 1/9

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Two brother X and Y appeared for an exam. The probability of selection of X is 1/7 and that of B is 2/9. Find the probability that both of them are selected.

A.

1/63

B.

1/14

C.

2/63

D.

1/9

Answer with explanation

Answer: Option CExplanation

Let A be the event that X is selected and B is the event that Y is selected.

P(A) = 1/7, P(B) = 2/9.

Let C be the event that both are selected.

P(C) = P(A) × P(B) as A and B are independent events:

= (1/7) × (2/9) = 2/63

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In a box, there are 8 red, 7 blue and 6 green balls. One ball is picked up randomly. What is the probability that it is neither red nor green?

A.

1/3

B.

3/5

C.

8/21

D.

7/21

Answer with explanation

Answer: Option AExplanation

Total number of balls = (8 + 7 + 6) = 21.

Let E = event that the ball drawn is neither red nor green

= event that the ball drawn is blue.

n(E) = 7.

P(E) = n(E)/n(S) = 7/21 = 1/3.

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A bag contains 4 red and 3 black balls. A second bag contains 2 red and 4 black balls. One bag is selected at random. From the selected bag, one ball is drawn. Find the probability that the ball drawn is red.

A.

23/42

B.

19/42

C.

7/32

D.

16/39

Answer with explanation

Answer: Option BExplanation

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A basket contains 10 apples and 20 oranges out of which 3 apples and 5 oranges are defective. If we choose two fruits at random, what is the probability that either both are oranges or both are non defective?

A.

136/345

B.

17/87

C.

316/435

D.

158/435

Answer with explanation

Answer: Option CExplanation

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A bag contains 6 black and 8 white balls. One ball is drawn at random. What is the probability that the ball drawn is white?

A.

3/7

B.

4/7

C.

1/8

D.

3/4

Answer with explanation

Answer: Option BExplanation

Let number of balls = (6 + 8) = 14.

Number of white balls = 8.

P (drawing a white ball) = 8 /14 = 4/7.

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In a race where 12 cars are running, the chance that car XX will win is 1/6, that YY will win is 1/10 and that ZZ will win is 1/8. Assuming that a dead heat is impossible. Find the chance that one of them will win

A.

47/120

B.

1/480

C.

1/160

D.

1/240

Answer with explanation

Answer: Option AExplanation

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A bag contains 7 green and 5 black balls. Three balls are drawn one after the other. The probability of all three balls being green, if the balls drawn are not replaced will be:

A.

343/1728

B.

21/36

C.

12/35

D.

7/44

Answer with explanation

Answer: Option DExplanation

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A box contains 100 balls, numbered from 1 to 100. If three balls are selected at random and with replacement from the box, what is the probability that the sum of the three numbers on the balls selected from the box will be odd?

A.

1/2

B.

3/4

C.

3/8

D.

1/8

Answer with explanation

Answer: Option AExplanation

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Four cards are drawn at random from a pack of 52 cards. What is the probability of getting all four cards of the same suit?

A.

2860/27725

B.

2560/270725

C.

2360/270725

D.

2860/270725

Answer with explanation

Answer: Option DExplanation

Total No. Of cards =52

No.Of cards for each suit = 13

Total types Of French suit = 4(hearts,spades, diamonds,clubs)

Prob. = Conditional case /total case

For this conditional case of selecting all four same suit,it can be either hearts or spades or clubs or diamonds

So Total no. Of conditional case= no. of case of selecting all four cards of hearts suit +no.of case of selecting all four cards of spades suit +no.of case of selecting all four cards of clubs suit + no. of case of selecting all four cards of diamonds suit.

= 13C4+13C4+13C4+ 13C4

=4*(13C4)=2860

Total no. of case= 52C4=270725

So, probability of getting all four cards of the same suit=2860/270725

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In a charity show tickets numbered consecutively from 101 through 350 are placed in a box.

What is the probability that a ticket selected at random (blindly) will have a number with a hundredth digit of 2?

A.

0.285

B.

0.40

C.

100/249

D.

99/250

Answer with explanation

Answer: Option BExplanation

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