Solution: Option (4) 27/35 Total number of balls = 4 pairs = 8 balls The total number of ways to select the four balls at random is 8C4 The... View Article
Solution: Option (2) 5/7 In a leap year, we have 366 days. So, we have 52 weeks and 2 days. So, the 2 days will the combination of {(Sunday,... View Article
Solution: Option (1) ½ Let X be the event of getting tails The required probability =P(X=1)+ P(x=3)+...+P(x=99) Using Binomial distribution,... View Article
Solution: Option (3) ⅓ The two girls seated together in 2! ways Let E be the event that 4 boys and 2 girls occupy the seat side by side. Thus,... View Article
Solution: Option (2) 5/54 When a die is rolled three times, the number of sample spaces obtained is 216. The number of favourable ways = 6C3... View Article
Solution: Option (2) ⅜ Total number of balls in a bag = 3+5 = 8 Number of white balls = 3 Therefore, the probability of selecting a white ball = ⅜... View Article
Solution: Option (1) 2/9 When two dice are thrown, the number of sample spaces obtained is 36. Favourable outcomes = {(1, 1) (2, 2) (4, 1) (1,... View Article
Solution: Option (4) (226 x 52C26)/104C26 In a deck, we have 52 cards. Therefore, 2 total number of cards in 2 decks = 52+ 52 = 104 cards... View Article
Solution: Option (4) 10/13 Total number of balls = 8+5 = 13 Favorable outcome= Event of selecting 2 red balls and 1 white ball + Event of... View Article
Solution: Option (1) 91/216 The required probability that the D4 shows a number appearing on one of D1, D2, and D3 is: = (6C1. ⅙. ⅙. ⅚. ⅚)3C1... View Article
Solution: Option (1) 2n/5n If the last of the number ends with 1, 3, 7 or 9, none of the numbers are even numbers. Hence, we have 4 choices,... View Article
Solution: Option (4) 2/7 Total number of horses = 7 Thus, the probability that Mr. x selects the winning horse = (1C1x 6C1)/7C2 = 6/21 = 2/7... View Article
Solution: Option (1) 35/102 Total number of students in class = 10+8 = 18 The probability of selecting 2 girls and 1 boy = (10C1x 8C2)/18C3... View Article
Solution: Option (2) 1/26 Total number of cards = 52 Total number of ace of heart = 1 Therefore, the required probability = (1C1×51C1)/52C2... View Article
Solution: Option (1) 7/744 In a chessboard, we have 64 squares Total Number of outcomes = 64C3 Total Number of favourable outcomes = 4(3C3 +... View Article
Solution: Option (2) 4/11 Total number of letters in “PROBABILITY” = 11 Number of vowels = 4 So, we can select 4 vowels at random. Hence, the... View Article
Solution: Option (3) ⅓ sum of numbers appearing on the upper face is observed to be 7) ={(1,6),(6,1),(2,5),(2,5),(5,2),(3,4),(4,3)} = 6... View Article
Solution: option (3) 2/3 Let g represent “a” girl and “b” represent a boy. A family has two children, out of which one kid is a boy. Hence,... View Article