Q1. A dice is thrown. Find the probability of getting an even number.

(A) 2/3

(B) 1

(C) 5/6

(D) 1/2

Answer:  (D)

Explanation: Total number of cases = 6 (1,2,3,4,5,6)

There are three even numbers 2,4,6

Therefore probability of getting an even number is:

P (even) = 3/6

⇒ P (even) = 1/2

Q2. Two coins are thrown at the same time. Find the probability of getting both heads.

(A) 3/4

(B) 1/4

(C) 1/2

(D) 0

Answer:  (B)

Explanation: Since two coins are tossed, therefore total number of cases = 2² = 4

Therefore, probability of getting heads in both coins is:

∴ P (head) = 1/4

Q3. Two dice are thrown simultaneously. The probability of getting a sum of 9 is:

(A) 1/10

(B) 3/10

(C) 1/9

(D) 4/9

Answer:  (C)

Explanation: Total cases = 36

Total cases in which sum of 9 can be obtained are:

(5, 4), (4, 5), (6, 3), (3, 6)

∴ P (9) = 4/36 = 1/9

Q4. 100 cards are numbered from 1 to 100. Find the probability of getting a prime number.

(A) 3/4

(B) 27/50

(C) 1/4

(D) 29/100

Answer:  (C)

Explanation: Total prime numbers from 1 to 100 are:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97

That means 25 out of 100

So probability is:

P (prime) = 25/100

⇒ P (prime) = 1/4

Q5. A bag contains 5 red balls and some blue balls .If the probability of drawing a blue ball is double that of a red ball, then the number of blue balls in a bag is:

(A) 5

(B) 10

(C) 15

(D) 20

Answer:  (B)

Explanation: Let the number of blue balls be x

Then total number of balls will be 5 + x.

According to question,

x/(5 + x) = 2 X (5/5+x)

⇒ x = 10

Q6.  A box of 600 bulbs contains 12 defective bulbs. One bulb is taken out at random from this box. Then the probability that it is non-defective bulb is:

(A) 143/150

(B) 147/150

(C) 1/25

(D) 1/50

Answer:  (B)

Explanation:

P (non-defective bulb) = 1 – P (Defective bulb)

= 1 – (12/600)

= (600 – 12)/600

= 588/600

= 147/150

Q7. Cards marked with numbers 2 to 101 are placed in a box and mixed thoroughly. One card is drawn from this box randomly, then the probability that the number on card is a perfect square.

(A) 9/100

(B) 1/10

(C) 3/10

(D) 19/100

Answer:  (B)

Explanation: The perfect square numbers between 2 to 101 are:

1, 4, 9, 16, 25, 36, 49, 64, 81, 100

Total numbers from 2 to 101 =100

So probability of getting a card with perfect square number is:

P (perfect square) = 10/100

⇒ P (perfect square) = 1/10

Q8. What is the probability of getting 53 Mondays in a leap year?

(A) 1/7

(B) 53/366

(C) 2/7

(D) 7/366

Answer:  (C)

Explanation: With 366 days, the number of weeks in a year is

366/7 = 52 (2/7)

i.e., 52 complete weeks which contains 52 Mondays,

Now 2 days of the year are remaining.

These two days can be

(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)

i.e., there are 7 pairs, in which Monday occurs in 2 pairs,

So probability is:

P (53 Monday) = 2/7

Q9. A card is drawn from a well shuffled deck of 52 cards. Find the probability of getting a king of red suit.

(A) 1/26

(B) 3/26

(C) 7/52

(D) 1/13

Answer:  (A)

Explanation: There are total 4 kings in 52 cards, 2 of red colour and 2 of black colour

Therefore, Probability of getting a king of red suit is:

P (King of red suit) = 2/52

⇒ P (King of red suit) = 1/26

Q10. A game of chance consists of spinning an arrow which is equally likely to come to rest pointing to one of the number 1,2,3……12 ,then the probability that it will point to an odd number is:

(A) 1/6

(B) 1/12

(C) 7/12

(D) 5/12

Answer:  (A)

Explanation: The odd numbers in 1,2,3……..12 are:

1,3,5,7,9,11

Therefore probability that an odd number will come is:

P (odd number) = 6/12

⇒ P (odd number) = 1/2

Q11. A game consists of tossing a one rupee coin 3 times and noting its outcome each time. Aryan wins if all the tosses give the same result i.e. three heads or three tails and loses otherwise. Then the probability that Aryan will lose the game.

(A) 3/4

(B) 1/2

(C) 1

(D) 1/4

Answer:  (A)

Explanation: Total outcomes are:

HHH, HHT, HTH, THH, HTT, THT, TTH, TTT

Favourable outcomes for losing game are

HHT, HTH, THH, HTT, THT, TTH

Therefore probability of losing the game is:

P (Losing the game) = 6/8

⇒ P (Losing the game) = 3/4

Q12. Riya and Kajal are friends. Probability that both will have the same birthday isthe same birthday is:

(A) 364/365

(B) 31/365

(C) 1/365

(D) 1/133225

Answer:  (C)

Explanation:

Riya may have any one of 365 days of the year as her birthday. Similarly Kajal may have any one of 365days as her birthday.

Total number of ways in which Riya and Kajal may have their birthday are:

365 × 365

Then Riya and Kajal may have same birthday on any one of 365 days.

Therefore number of ways in which Riya and Kajal may have same birthday are:

= 365/365 X 365

= 1/365

Q13. A number x is chosen at random from the numbers -2, -1, 0 , 1, 2. Then the probability that x² < 2 is?

(A) 1/5

(B) 2/5

(C) 3/5

(D) 4/5

Answer:  (C)

Explanation: We have 5 numbers −2,−1,0,1,2

Whose squares are 4,1,0,1,4

So square of 3 numbers is less than 2

Therefore Probability is:

P (x² < 2) = 3/5

Q14. A jar contains 24 marbles. Some are red and others are white. If a marble is drawn at random from the jar, the probability that it is red is 2/3, then the number of white marbles in the jar is:

(A) 10

(B) 6

(C) 8

(D) 7

Answer:  (C)

Explanation: Let the number of white marbles be x.

Since only two colour marbles are present, and total probability we know of all the events is equal to 1.

P (white) = 1 – P (red)

x/24 = 1 – (2/3)

⇒ x/24 = 1/3

⇒ x = 8

So there are 8 white marbles.

Q15. A number is selected at random from first 50 natural numbers. Then the probability that it is a multiple of 3 and 4 is:

(A) 7/50

(B) 4/25

(C) 1/25

(D) 2/25

Answer:  (D)

Explanation: The numbers that are multiple of 3(from first 50 natural numbers) are:

3, 6, 9, 12, 15, 18………………..48

The numbers that are multiple of 4 (from first 50 natural numbers) are:

4, 8, 12, 16…………………….48

The numbers that are multiples of 3 and 4 both are the multiples of 3×4=12 as both 3 and 4 are co-prime.

So common multiples are:

12, 24, 36, 48

Therefore probability is:

P (multiple of 3 and 4) = 4/50

⇒ P (multiple of 3 and 4) = 2/25

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