## Probability MCQ1) An event in the probability that will never be happened is called as - - Unsure event
- Sure event
- Possible event
- Impossible event
2) What will be the value of P(not E) if P(E) = 0.07? - 90
- 007
- 93
- 72
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 3) What will be the probability of getting odd numbers if a dice is thrown? - 1/2
- 2
- 4/2
- 5/2
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 4) What is the probability of getting a sum as 3 if a dice is thrown? - 2/18
- 1/18
- 4
- 1/36
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 5) What is the probability of getting an even number when a dice is thrown? - 1/6
- 1/2
- 1/3
- 1/4
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 6) The probability of getting two tails when two coins are tossed is - - 1/6
- 1/2
- 1/3
- 1/4
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 7) What is the probability of getting the sum as a prime number if two dice are thrown? - 5/24
- 5/12
- 5/30
- 1/4
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 8) What is the probability of getting atleast one head if three unbiased coins are tossed? - 7/8
- 1/2
- 5/8
- 8/9
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 9) What is the probability of getting 1 and 5 if a dice is thrown once? - 1/6
- 1/3
- 2/3
- 8/9
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 10) What will be the probability of losing a game if the winning probability is 0.3? - 0.5
- 0.6
- 0.7
- 0.8
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 11) If two dice are thrown together, what is the probability of getting an even number on one dice and an odd number on the other dice? - 1/4
- 3/5
- 3/4
- 1/2
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 12) In a box, there are 8 orange, 7 white, and 6 blue balls. If a ball is picked up randomly, what is the probability that it is neither orange nor blue? - 1/3
- 1/21
- 2/21
- 5/21
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 13) A card is drawn from a pack of 52 cards. What is the probability of getting a king of a black suit? - 1/26
- 1/52
- 3/26
- 7/52
The event of getting a black king = 2 So, n (E) = 2 P(E) = n(E) / n(S) = 2 / 52 = 1 / 26 14) A dice is thrown twice. What is the probability of getting two numbers whose product is even? - 6/4
- 1/2
- 5/4
- 3/4
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 15) Suppose a number x is chosen from the numbers -2, -1, 0, 1, 2. What will be the probability of x - 1/5
- 2/3
- 3/5
- 4/5
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 x 16) If a number is selected at random from the first 50 natural numbers, what will be the probability that the selected number is a multiple of 3 and 4? - 7/50
- 4/25
- 2/25
- None of the above
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 17) What is the probability of getting a prime number from the numbers started from 1 to 100? - 1/4
- 1/100
- 1/25
- None of the above
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 18) What is the probability of drawing an ace from a pack of 52 cards? - 4/13
- 1/13
- 1/52
- None of the above
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 19) In 30 balls, a batsman hits the boundaries 6 times. What will be the probability that he did not hit the boundaries? - 1/5
- 4/5
- 3/5
- None of the above
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 20) Which of the following probability cannot exist? - 2/5
- -1.5
- 7
- None of the above
21) A card is drawn from a pack of 52 cards. What is the probability of getting a queen card? - 1/26
- 1/52
- 3/13
- 1/13
Total number of queen cards = 4 The probability of getting a queen card is = 4/52 or 1/13 22) What will be the probability of an impossible event? - 0
- 1
- Infinity
- None of the above
23) Which of the following can be the probability of an event? - -1.3
- 004
- 3/8
- 10/7
24) If three coins are tossed simultaneously, what is the probability of getting two heads together? - 3/8
- 1/8
- 5/8
- None of the above
Let E is the event of getting two heads together Then E = {THH, HHT, HHH} P(E) = n(E) / n(S) = 3/8. 25) The probability of winning the first prize in a lottery of a girl is 8/100. If the total of 6000 tickets are sold, then how many tickets the girl purchased? - 480
- 750
- 280
- None of the above
Total number of tickets sold = 6000 No. of tickets girl purchased = 8/100 * 6000 = 480 26) There are 3 blue socks, 5 brown socks, and 4 white socks in a drawer. If two socks are picked up randomly, what is the probability that the selected socks are of the same color? - 1
- 0
- 19/66
- 4/11
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 27) A card is drawn from a pack of 52 cards. What is the probability that it is a face card (King, Queen, and Jack only)? - 1/26
- 2/13
- 1/13
- 3/13
No. of face cards = 12 Probability of getting a face card = 12/52 or 3/13 28) A stock of pens consists of 144 ball pens in which 20 pens are defective, and others are good. A girl went to the shop to purchase a pen. The shopkeeper randomly draws one pen and gives it to her. What is the probability that a girl will buy the good pen? - 5/26
- 5/36
- 31/36
- None of the above
No. of defective pens = 20 No. of good pens = 124 The probability of buying a good pen is = 124/144 = 31/36. 29) The probability of randomly selecting a rotten apple is 0.18 from the heap of 900 apples. So, what is the number of rotten apples in a heap? - 162
- 164
- 136
- 160
Probability of selecting a rotten people = 0.18 Total number of rotten apples in a heap = 900 * 0.18 = 162. 30) If a number is selected at random from the first 100 natural numbers, what will be the probability that the selected number is a perfect cube? - 1/25
- 2/25
- 3/25
- 4/25
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 31) What will be the number of events if 10 coins are tossed simultaneously? - 512
- 90
- 1000
- 1024
32) If two dice are thrown simultaneously, what is the probability of getting a multiple of 2 on one dice and multiple of 3 on the other dice? - 5/4
- 5/12
- 11/36
- 1/2
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 33) Four dice are thrown simultaneously. What will be the probability that all of them have the same face? - 1/6
- 1/36
- 1/216
- None of the above
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. 34) Two people X and Y apply for a job in a company. The probability of the selection of X is 2/5, and Y is 4/7. What is the probability that both of them get selected? - 1/6
- 27/35
- 8/35
- 3/35
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 35) Two dice are thrown simultaneously. What will be the probability of getting a sum of 7? - 1/6
- 2/9
- 5/6
- None of the above
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 36) A card is drawn from a pack of 52 cards. What is the probability of getting a non-face card? - 10/13
- 3/13
- 1/13
- None of the above
No. of face cards = 12 Non-face cards = 52 - 12 = 40 Probability of getting a non-face card = 40/52 or 10/13 37) A school has five houses named as A, B, C, D, and E. There are 23 students in a class in which 4 students are from house A, 8 students are from house B, 5 from C, 2 from D, and the rest from house E. Class teacher randomly selects a student to be the class monitor. What is the probability that the selected student is not from house A, B, and C? - 1/23
- 2/23
- 5/23
- 6/23
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 38) A dice is rolled. The probability of getting a number x where 1 ? x ? 6 is - - Greater than 0
- Greater than 1
- Between 1 and 0
- Equal to 1
P (sure outcome) = 1 {1, 2, 3, 4, 5 6} is called sure event 39) If three coins are tossed simultaneously, what is the probability of getting atmost two heads? - 7/8
- 1/8
- 5/8
- None of the above
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. 40) Which of the following statement is not true about probability? - The probability of an impossible event is 0.
- Probability can be greater than 1 or less than 0.
- Probability cannot be greater than 1.
- None of the above
41) If P is the probability of an event, what is the probability of its complement? - 1 - 1/P
- P - 1
- 1 - P
- None of the above
42) The probability of selecting a bad egg is 0.035 from the lot of 400 eggs. So, what is the number of bad eggs in the lot? - 14
- 16
- 18
- 20
Probability of selecting a bad egg = 0.035 Total number of bad eggs in the lot = 400 * 0.035 = 14. 43) Using the digits 1, 2, 3, 4, and 5, a number having five digits is formed without any repetition. What is the probability that the number is divisible by 4? - 1/5
- 2/5
- 5/6
- 3/5
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. 44) The set of one or more than one outcomes from an experiment is called as - - Z-value
- Arithmetic mean
- Event
- None of the above
45) If one event occurs, another event cannot happen, i.e., the events that cannot occur simultaneously are called as - - Exhaustive Events
- Mutually exclusive events
- Equally likely events
- Independent events
46) What is the probability of the random arrangement of letters in the word "UNIVERSITY" and two I's should come together? - 1/7
- 3/5
- 1/5
- 2/7
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 47) In class, 30% of students study Hindi, 45% study Maths, and 15% study both Hindi and Maths. If a student is randomly selected, what is the probability that he/she study Hindi or maths? - 1/5
- 3/5
- 2/5
- 2/7
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 48) In binomial distribution, successive trials are - - Mutually exclusive
- Dependent
- Independent
- None of the above
49) The formula for finding the mean of the binomial distribution is - - np
- (1 - p)
- n + p
- None of the above
50) How many parameters in the binomial distribution? - 1
- 2
- 3
- 4
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