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A bag contains 6 red balls 11 yellow balls and 5 pink balls. If two balls are drawn at random from the bag. One after another what is the probability that the first ball is red and second ball is yellow?

A.

2/7

B.

1/14

C.

3/14

D.

5/7

Answer with explanation

Answer: Option AExplanation

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One card is drawn at random from a pack of 52 cards. What is the probability that the card drawn is a face card (Jack, Queen and King only)?

A.

(1/13)

B.

(3/13)

C.

(1/4)

D.

(9/52)

Answer with explanation

Answer: Option BExplanation

Clearly, there are 52 cards, out of which there are 12 face cards.

Probability of getting a face card = 12/52 = 3/13

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Two friends A and B apply for a job in the same company. The chances of A getting selected is 2/5 and that of B is 4/7. What is the probability that both of them get selected?

A.

8/35

B.

34/35

C.

27/35

D.

None of these

Answer with explanation

Answer: Option AExplanation

P(A) = 2/5

P(B) = 4/7

E = {A and B both get selected}

P(E) = P(A)*P(B)

= 2/5 * 4/7

= 8/35

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The probability that Soumya will get marry within 365 days is ‘a’ and the probability that her colleague Alma get marry within 365 days is ‘b’. Find the probability that only one of the two gets marry at the end of 365 days.

A.

a-b-2ab

B.

a+b-2ab

C.

a-b+2ab

D.

ab-a-b

Answer with explanation

Answer: Option BExplanation

The probability that Soumya will get marry than Alma will be =a(1-b).

Similarly, The probability that Alma will get marry than Soumya will be b(1-a).

The probability thateither of these two get marry =a(1-b) +b(1-a) =a+b-2ab.

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10 books are placed at random in a shelf. The probability that a pair of books will always be together is -.

A.

1/10

B.

9/10

C.

1/5

D.

1/10

Answer with explanation

Answer: Option CExplanation

10 books can be rearranged in 10! ways consider the two books taken as a pair then number of favourable ways of getting these two books together is 9! 2!

Required probability = 1/5

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There are 8 boys and 12 girls in a class. 5 students have to be chosen for an educational trip. Find the number of ways in which this can be done if 2 particular girls are always included.

A.

805

B.

816

C.

961

D.

1050

Answer with explanation

Answer: Option BExplanation

18c3 = 816 (2 girls already selected).

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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.

1/8

B.

5/9

C.

4/7

D.

3/8

Answer with explanation

Answer: Option CExplanation

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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A box contains nine bulbs out of which 4 are defective. If four bulbs are chosen at random, find the probability that atleast one bulb is good.

A.

6/63

B.

2/63

C.

125/126

D.

1/126

Answer with explanation

Answer: Option CExplanation

Required probability = 1 – 1/126 = 125/126

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Determine the probability that a digit chosen at random from the digits 1, 2, 3, …12 will be odd.

A.

1/2

B.

4/9

C.

1/9

D.

5/9

Answer with explanation

Answer: Option AExplanation

Total no. of Digits = 12. Equally likely cases = 12.

There are six odd digits. Probability = 6 / 12 = 1 / 2

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There are 7 red balls and 8 yellow balls in a bag. Two balls are simultaneously drawn at random. What is the probability that both the balls are of same colour ?

A.

B.

C.

D.

Answer with explanation

Answer: Option DExplanation

Total balls in the bag = 7 + 8 = 15

Total possible outcomes = Selection of 2 balls out of 15 balls

= 15C_{2} = (15*14) / (1*2) = 105

Total favourable outcomes = Selection of 2 balls out of 8 yellow balls + Selection of 2 balls out of 7 red balls

= 8C_{2} + 7C_{2} = [(8*7)/(1*2)] + [(7*6)/(1*2)]

= 28 + 21 = 49

Required probability = 49/105 = 7/15

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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.

3/4

B.

1/4

C.

4/7

D.

3/8

Answer with explanation

Answer: Option DExplanation

Total no.on ways to stay in a hotel without any condition = 4 x 4 x 4 = 64 ways.

Bcoz, every person have 4 choices

Now, if every person has to stay in the different hotels then no.of ways = 4×3×2 = 24 ways.

Required probability =24/64 = 3/8

(or)

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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From a pack of cards two cards are drawn one after the other, with replacement. The probability that the first is a red card and the second is a king is -.

A.

3/52

B.

1/26

C.

11/26

D.

15/26

Answer with explanation

Answer: Option BExplanation

Let E_{1} be the event of drawing a red card.

Let E_{2} be the event of drawing a king .

P(E_{1} ∩ E_{2}) = P(E_{1}) . P(E_{2})

(As E_{1} and E_{2} are independent)

= 1/2 * 1/13 = 1/26

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A.

1/10

B.

1/5

C.

9/10

D.

6/5

Answer with explanation

Answer: Option BExplanation

10 books can be rearranged in 10! ways consider the two books taken as a pair then number of favourable ways of getting these two books together is 9! 2!

Required probability = 1/5

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A.

6/63

B.

2/63

C.

125/126

D.

1/126

Answer with explanation

Answer: Option CExplanation

Required probability = 1 – 1/126 = 125/126

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A.

4/7

B.

3/8

C.

5/9

D.

1/6

Answer with explanation

Answer: Option AExplanation

Explanation: 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 X will win is 1/6, that will win is 1/10 and that Z 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.

1/180

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.

7/44

B.

12/35

C.

21/36

D.

343/1728

Answer with explanation

Answer: Option AExplanation

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What is the probability of getting a sum 9 from two throws of dice.

A.

1/3

B.

1/9

C.

1/12

D.

2/9

Answer with explanation

Answer: Option BExplanation

Total number of cases = 6*6 = 36

Favoured cases = [(3,6), (4,5), (6,3), (5,4)] = 4

So probability = 4/36 = 1/9

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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 party there are 5 couples. Out of them 5 people are chosen at random. Find the probability that there are at the least two couples ?

A.

6/7

B.

19/21

C.

7/31

D.

5/21

Answer with explanation

Answer: Option CExplanation

Number of ways of (selecting at least two couples among five people selected) = (⁵C₂ x ⁶C₁)

As remaining person can be any one among three couples left.

Required probability = (⁵C₂ x ⁶C₁)/¹⁰C₅

= (10 x 6)/252 = 5/21

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There are 10 Letters and 10 correspondingly 10 different Address. If the letter are put into envelope randomly, then find the Probability that Exactly 9 letters will at the Correct Address ?

A.

1/10

B.

1/9

C.

1

D.

0

Answer with explanation

Answer: Option DExplanation

As we know we have 10 letter and 10 different address and one more information given that exactly 9 letter will at the correct address….so the remaining one letter automatically reach to their correct address

P(E) = favorable outcomes /total outcomes

Here favorable outcomes are ‘0’.

So probability is ‘0’.

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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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