Probability – Aptitude Questions and Answers | Latest Online Probability MCQ Aptitude Test
Probability – Aptitude Questions and Answers | Online Probability MCQ Aptitude Test Quiz Name Probability Category Online Aptitude Test Number of Questions 15 Time 30 Minutes... explore below more
Probability – Aptitude Questions and Answers | Online Probability MCQ Aptitude Test
Quiz Name
Probability
Category
Online Aptitude Test
Number of Questions
15
Time
30 Minutes
Exam Type
MCQ (Multiple Choice Questions)
1. Tickets with numbers ranging from 1 to 20 are mingled together and a ticket is selected at random. What is the chance that the ticket drawn has a number that is a multiple of three or five? a. 1/2 b.2/5 c. 8/15 d. 9/20
Answer: D
Explanation: S = 1, 2, 3, 4,…., 19, 20 in this case.E = occurrence of a multiple of 3 or 5 = 3, 6, 9, 12, 15, 18, 5, 10, 20.P(E) = n(E)/n(S) = 9/20 P(E) .
2. There are two red, three green, and two blue balls in each bag. At random, two balls are selected. What are the chances that none of the balls drawn will be blue? a. 10/21 b. 11/21 c. 2/7 d. 5/7
Answer: A
Explanation: Total number of balls = (7 + 2 + 3 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 +….)Assume that S is the sample space.Then, n(S) = Number of ways to pull two balls from a set of seven = 7C2 = (7 x 6)/ (2 x 1),Let E stand for the event of drawing two balls, none of which are blue.n(E) is the number of different methods to draw two balls from a set of (2 + 3) balls.5C2 = (5 x 4)/ = 5C2 = (5 x 4)/ = 5C2 = (5 x (2 x 1),P(E) = n(E)/n(S) = 10/21.
3. There are 8 red, 7 blue, and 6 green balls in a box. A single ball is chosen at random. How likely is it that it is neither red nor green? a. 1/3 b. 3/4 c.7/19 d. 8/21
Answer: A
Explanation: The total number of balls is 21 (8 + 7 + 6).Let E denote the possibility that the drawn ball is neither red nor green, and B denote the possibility that the drawn ball is blue.n(E) = n(E) = n(E) = P(E) = n(E)/n(S) = 7/21 = ⅓
4. What is the chance of achieving a total of 9 from two dice throws? a. 1/6 b. 1/8 c. 1/9 d. 1/12
Answer: C
Explanation: n(S) = (6 x 6) = 36 in two dice rolls.Let E denote the occurrence of obtaining a total of (3, 6), (4, 5), (5, 4), (6, 3).P(E) = n(E)/n(S) = 4/36 = 1/9.
5. Three equally weighted coins are tossed. What are the chances of receiving at least two heads? a. 3/4 b. 1/4 c. 3/8 d. 7/8
Answer: D
Explanation: S = TTT, TTH, THT, HTT, THH, HTH, HHT, HHH, THH, THH, THH, THH, THH, THH, THH, THH, THH, THH, THH, THH, THH,Let E denote the occurrence of at least two heads.Then E = TTT, TTH, THT, HTT, THH, HTH, HHT, HTT, THH, HTH, HHT.n(E)/n(S) = 7/8 P(E) = n(E)/n(S)
6. Two dice are thrown at the same time. What is the likelihood of receiving two numbers with an even product? a. 1/2 b. 3/4 c. 3/8 d. 5/16
Answer: B
Explanation: We get n(S) = (6 x 6) = 36 when we toss two dice at the same time.Then E = (1, 2), (1, 4), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 2), (3, 4), (4, 5), (4, 6), (5, 2), (5, 4), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6), (6, 6),P(E) = n(E)/n(S) = 27/36 = ¾.
7. There are 15 males and 10 girls in each class. At random, three pupils are chosen. The likelihood of one girl and two boys being chosen is: a. 21/46 b. 25/117 c. 1/50 d. 3/25
Answer: A
Explanation: Let S represent the sample space, and E represent the event of picking one female and two males.Then, n(S) = Number of methods to choose three students from a group of 25 = 25C3 = (25 x 24 x 23)/ (3 x 2 x 1),n(E) = (10C1 x 15C2) = 10 x (15 x 14) / 2300 (2 x 1).P(E) = n(E)/n(S) = 1050/2300 = 21/46
8. There are ten prizes and twenty-five blanks in a lottery. The winner of a lottery is chosen at random. What are the chances of winning a prize? a. 1/10 b. 2/5 c. 2/7 d. 5/7
Answer: C
Explanation: P = 10/(10 + 25) = 10/35 = 2/7 (winning a reward).
9. Two cards are selected at random from a deck of 52 cards. What are the chances that both cards are kings? a. 1/15 b. 25/57 c. 35/256 d.1/221
Answer: D
Explanation: Assume that S is the sample space.As a result, n(S) = 52C2 = (52 x 51) /(2 x 1) = 1326.Let E represent the probability of receiving two kings out of four.(4 x 3)/(2 x 1) = 6. n(E) = 4C2 = (4 x 3)/(2 x 1) = 6.P(E) = n(E)/n(S) = 6/1326 = 1/221 .
10. A pair of dice is tossed. The chances of the total score being a prime number are: a. 1/6 b. 5/12 c. 1/2 d. 7/9
Answer: B
Explanation: n(S) clearly equals (6 x 6) = 36.Let E be the probability that the total 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),P(E) = n(E)/n(S) = 15/36 = 5/12 .
11. A card is chosen at random from a deck of 52 cards. The chances of receiving a queen of clubs or a king of hearts are as follows: a. 1/13 b. 2/13 c. 1/26 d. 1/52
Answer: C
Explanation: n(S) = 52 in this case.Let E denote the possibility of receiving a queen of clubs or a king of hearts.n(E) = 2 then.P(E) = n(E)/n(S) = 2/52 = 1/26 .
12. There are four white, five red, and six blue balls in each bag. Three balls are randomly selected from the bag. The likelihood that they’re all red is: a. 1/22 b. 3/22 c. 2/91 d. 2/27
Answer: C
Explanation: Assume that S is the sample space.Then, n(S) = number of ways to pull three balls out of 15 = 15C3 = (15 x 14 x 13)/ (3 x 2 x 1)= 455, where E denotes the event of obtaining all three red balls.5C3 = 5C2 = (5 x 4)/(2 x 1) = 10. n(E) = 5C3 = 5C2 = (5 x 4)/(2 x 1) = 10.P(E) = n(E)/n(S) = 10/455 = 2/91 .
13. Two cards are chosen at random from a deck of 52. The chances of one being a spade and the other being a heart are: a. 3/20 b. 29/34 c. 47/100 d. <13 br="">
Answer: D
Explanation: Assume that S is the sample space.As a result, n(S) = 52C2 = (52 x 51)/(2 x 1) = 1326.Let E stand for the event of receiving one spade and one heart.P(E) = n(E)/n(S) = 169/1326 = 13/102. n(E) = number of ways to choose 1 spade out of 13 and 1 heart out of 13 = (13C1 x 13C1) = (13 x 13) = 169.
14. From a pack of 52 cards, one card is chosen at random. What is the chance that the card drawn will be a face card (only Jack, Queen, and King)? a. 1/13 b. 3/13 c. 1/4 d. 9/52
Answer: B
Explanation: Clearly, there are 52 cards in all, 12 of which are face cards.P = 12/52 = 3/13 (receiving a face card).
15. There are 6 black and 8 white balls in each bag. A single ball is chosen at random. What is the chance that the drawn ball is white? a. 3/4 b. 4/7 c. 1/8 d. 3/7
Answer: B
Explanation: Let’s say the total number of balls is 6 + 8 = 14.The number of white balls is eight.P = 8/14 = 4/7 (drawing a white ball).
