Answer: (d) Impossible event

Explanation: An event that will never be happened is known as the impossible event. For example – Tossing double-headed coins and getting tails in an impossible event, rolling a die and getting number > 10 in an impossible outcome, etc.

Answer: (c) 0.93

Explanation: If the probability of happening of an event P(E) and that of not happening is P(E), then

P(E) + P(not E) = 1

So, P(not E) = 1 – P(E)

Since P(E) = 0.07

P(not E) = 1 – 0.07

P(not E) = 0.93

Answer: (a) 1/2

Explanation: The sample space when a dice is rolled, S = (1, 2, 3, 4, 5, and 6)

So, n (S) = 6

E is the event of getting an odd number.

So, n (E) = 3

Probability of getting an odd number P (E) = Total number of favorable outcomes / Total number of outcomes

n(E) / n(S) = 3/6 = 1/2

Answer: (b) 1/18

Explanation: In two throws a dice, n (S) = 6 * 6 = 36

Let E is the event of getting a sum of three.

E = (1, 2), (2, 1)

So, n (E) = 2

So, P (E) = n(E) / n(S) = 2/36 or 1/18

Answer: (b) 1/2

Explanation: The sample space when a dice is rolled, S = (1, 2, 3, 4, 5, and 6)

So, n (S) = 6

E is the event of getting an even number.

So, n (E) = 3

Probability of getting an even number P (E) = Total number of favorable outcomes/Total number of outcomes

n(E) / n(S) = 3/6 = 1/2

Answer: (d) 1/4

Explanation: The sample space when two coins are tossed = (H, H), (H, T), (T, H), (T, T)

So, n(S) = 4

The event “E” of getting two tails (T, T) = 1

So, n(E) = 1

So, the probability of getting two tails, P (E) = n(E) / n(S) = 1/4

Answer: (b) 5/12

Explanation: As per the question: n (S) = 6*6 = 36

And, the event that the sum is a prime number:

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)}

So, n (E) = 15

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

Answer: (a) 7/8

Explanation: The sample space is = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}

Let E is the event of getting atleast one head

Then E = {TTH, THT, HTT, THH, HTH, HHT, HHH}

P(E) = n(E) / n(S) = 7/8

Answer: (b) 1/3

Explanation: The sample space when a dice is rolled, S = (1, 2, 3, 4, 5 and 6)

So, n (S) = 6

E is the event of getting 1 and 5

So, n (E) = 2

P (E) = Total number of favorable outcomes / Total number of outcomes

n(E) / n(S) = 2/6 = 1/3

Answer: (c) 0.7

Explanation: Let P(E) is the probability of winning the game, and P(not E) be the probability of not winning the game.

P(E) + P(not E) = 1

So, P(not E) = 1 – P(E)

Since P(E) = 0.3

P(not E) = 1 – 0.3

P(not E) = 0.7

Answer: (d) 1/2

Explanation: As per the question: n (S) = 6*6 = 36

Let E be the event of getting an even number on one die and an odd number on the other

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

So, n (E) = 18

n(E) / n(S) = 18/36 = 1/2

Answer: (a) 1/3

Explanation: Total number of balls or sample space = 8 + 7 + 6 = 21

So, n(S) = 21

Let E is the event that the ball drawn is neither orange nor blue or event that the drawn ball is white. There are 7 white balls.

So, n(E) = 7

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

Answer: (a) 1/26

Explanation: We have, n (S) = 52

The event of getting a black king = 2

So, n (E) = 2

P(E) = n(E) / n(S) = 2 / 52 = 1 / 26

Answer: (d) 3/4

Explanation: In a simultaneous throw of the two dice, the sample space, S = 6 * 6 = 36

So, n (S) = 36

The event “E” = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

So, n (E) = 27

P(E) = n(E)/n(S) = 27/36 = 3/4

Answer: (d) 4/5

Explanation: The numbers given in the question are -2, -1, 0, 1, 2.

The squares of these numbers are 4, 1, 0, 1, 4. So the square of four numbers is greater than 0.

Therefore, the probability of x² > 0 is 4/5.

Answer: (c) 2/25

Explanation: We have first 50 natural numbers.

There are four common multiples of 3 and 4 from the first 50 natural numbers that are = 12, 24, 36, 48

So, P(multiple of 3 and 4) = 4/50 or 2/25

Answer: (a) 1/4

Explanation: We have given the first 100 natural numbers.

There are twenty-five prime numbers from the first 100 natural numbers that are = 2, 3, 5, 7, 9, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97

So, the probability of prime numbers from 1 to 100 is

P(prime) = 25/100 or 1/4

Answer: (b) 1/13

Explanation: We have, n (S) = 52

There are 4 aces in a deck of card, so the probability of drawing an ace from a deck of card is: 4/52 = 1/13

Answer: (b) 4/5

Explanation: Total number of balls = 30

No. of boundaries the batsman hit = 6

No. of balls without boundaries = 30 – 6 = 24

So, the probability when there is no boundary = 24/30 = 4/5

Answer: (b) -1.5

Explanation: Probability cannot be negative as it lies between 0 and 1.

Answer: (d) 1/13

Explanation: We have, total number of cards = 52

Total number of queen cards = 4

The probability of getting a queen card is = 4/52 or 1/13

Answer: (a) 0

Explanation: An event that will never be happened is known as the impossible event.The probability of an impossible event is 0.

Answer: (c) 3/8

Explanation: The probability of an event neither exceeds unity nor can it be negative. It lies between 0 and 1.

Answer: (a) 3/8

Explanation: The sample space is = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}

Let E is the event of getting two heads together

Then E = {THH, HHT, HHH}

P(E) = n(E) / n(S) = 3/8.

Answer: (a) 480

Explanation: Probability of winning the first prize of the girl = 8/100

Total number of tickets sold = 6000

No. of tickets girl purchased = 8/100 * 6000 = 480

Answer: (c) 19/66

Explanation: Total number of socks = 3 + 5 + 4 = 12 socks

Probability of selecting first blue sock = 3/12 or 1/4

Probability of selecting second blue sock = 2/11

Probability of selecting two blue socks = 1/4 * 2/11 = 2/44 or 1/22

Similarly, the probability of selecting two brown socks = 5/33

Similarly, the probability of selecting two white socks = 1/11

So, the probability of having two socks with the same color = probability of having two blue socks + probability of having two brown socks + probability of having two white socks

= 1/22 + 5/33 + 1/11

= 19/66

Answer: (d) 3/13

Explanation: Total number of cards = 52

No. of face cards = 12

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

Answer: (c) 31/36

Explanation: Total number of ball pens = 144

No. of defective pens = 20

No. of good pens = 124

The probability of buying a good pen is = 124/144 = 31/36.

Answer: (a) 162

Explanation: Total number of apples in a heap = 900

Probability of selecting a rotten people = 0.18

Total number of rotten apples in a heap = 900 * 0.18 = 162.

Answer: (a) 1/25

Explanation: We have the set of 100 natural numbers

There are four perfect cubes from the first 100 natural numbers that are = 1, 8, 27, and 64

So, the probability of selecting a perfect cube is = 4/100 or 1/25

Answer: (d) 1024

Explanation: None

Answer: (c) 11/36

Explanation: Let S be the sample space, and E be the required events

E = {(2, 3), (3, 2), (2, 6), (6, 2), (4, 3), (3, 4), (4, 6), (6, 4), (3, 6), (6, 3), (6, 6)}

n(E) = 11 and n(S) = 36

So, the probability will be = n(E)/n(S) = 11/36

Answer: (c) 1/216

Explanation: The total number of elementary events = 6⁴ = 1296

So, n(S) = 1296

Let E be the event that all dice show the same face

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

n(E) = 6

So, the probability will be = n(E)/n(S) = 6/1296 or 1/216.

Answer: (c) 8/35

Explanation: P(X) = 2/5

P(Y) = 4/7

Let E be the event that both of them get selected

So, P(E) = P(X) * P(Y)

= 2/5 * 4/7

= 8/35

Answer: (a) 1/6

Explanation: The sample space n(S) = 6 * 6 = 36

Let E be the event that the sum is 7

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

n(E) = 6

P(E) = 6/36 = 1/6

Answer: (a) 10/13

Explanation: The sample space n(S) = 50 cards

No. of face cards = 12

Non-face cards = 52 – 12 = 40

Probability of getting a non-face card = 40/52 or 10/13

Answer: (d) 6/23

Explanation: Total number of students = 23

No. of students in houses A, B, and C = 4 + 8 + 5 = 17

Let E be the remaining students

So, no. of remaining students n(E) = 23 – 17 = 6

Probability that selected student is not from house A, B, and C = 6/23

Answer: (d) Equal to 1

Explanation: An Outcome that will definitely happen is a sure outcome. Rolling a die and getting a number that is greater than equal to 1 and less than equal to 6 is a sure outcome.

P (sure outcome) = 1

{1, 2, 3, 4, 5 6} is called sure event

Answer: (a) 7/8

Explanation: The sample space is = {TTT, TTH, THT, HTT, THH, HTH, HHT, HHH}

Let E is the event of getting atmost two heads

Then E = {TTT, TTH, THT, HTT, THH, HTH, HHT}

n(E) = 7

P(E) = n(E) / n(S) = 7/8.

Answer: (b) Probability can be greater than 1 or less than 0.

Explanation: The probability of an event neither exceeds unity nor can it be negative. It lies between 0 and 1.

Answer: (c) 1 – P

Explanation: None

Answer: (a) 14

Explanation: Total number of eggs in the lot = 400

Probability of selecting a bad egg = 0.035

Total number of bad eggs in the lot = 400 * 0.035 = 14.

Answer: (a) 1/5

Explanation: If a number is formed using the digits 1, 2, 3, 4, and 5 and if it should be divisible by 4, then the last two digits of the number should be 12 or 24, 32 or 52.

So, the number can be formed using the remaining three digits, i.e., 3! = 6 ways.

The number divisible by 4 using the given digits can be formed by = 6 * 4 = 24 ways.

Total numbers formed using the given digits = 5! = 120

So, the required probability = 24/120 = 1/5.

Answer: (c) Event

Explanation: The performance of an experiment is called a trial, and the set of its outcomes is termed an event.

Answer: (b) Mutually exclusive events

Explanation: Events are called mutually exclusive if they cannot occur simultaneously.

Answer: (c) 1/5

Explanation: The total number of words that can be formed using the letters of the word “UNIVERSITY” and the two I’s should come together is = 10!/2!

If we consider two I’s as one letter, the number of ways of arrangement in which both I’s are together = 9!

The required probability is = 9!/10!/2! = 2/10 = 1/5

Answer: (b) 3/5

Explanation: Given that 45% study Hindi, i.e., P(H) = 45/100 = 9/20

30% study Maths, i.e., P(M) = 30/100 = 6/20

15% study both Hindi and Maths, i.e., P(H and M) = 15/100 = 3/20

So, P(H or M) = P(H) + P(M) – P(H and M)

= 9/20 + 6/20 – 3/20

= 12/20 or 3/5

Answer: (c) Independent

Explanation: None

Answer: (a) np

Explanation: None

Answer: (b) 2

Explanation: None