# Question Bank - Probability

• Q19) If two squares are choosen from a chess board with 64 squares; what is the probability that they share a common side?

• Q20) What is the probability that a point satisfying the conditions |x| ≤ 2 and |y| ≤ 2 also satisfies |x + y| ≤ 1?

• Q21) There are four envelops with letters. Two are chosen at random and opened and are found wrongly addressed. Find the probability that there are exactly two wrongly addressed envelops

• Q22) If two natural numbers, each of which is a 2-digit, are chosen randomly, find the probability that these two numbers have the same unit digits

• Q23) Two different prime numbers less than 100 are randomly selected and multiplied . What is the probability that product is divisible by 9

• Q24) If 60% of humans are vegetarian.What is the probability that you are a vegetarian?

• Q25) There are 3 concentric circular strips on a dart. The probability of hitting the innermost lamina us 1/9,that of central strip is 1/3 and of the outermost strip is 5/9. One gets 10 points for hitting the central most lamina,6 for central strip and 2 for the outermost strip. What is the probability of getting atleast 20 points in 3 attempts? Given that a target is never missed?

• Q26) A Coin is tossed 5 times, then the probability that same face does not show up on any 3 consequetive flips is?

• Q27) If three numbers are chosen in the set containing from 1 to 15, find the probability that their sum is divisible by 3

• Q28) There are 2n consecutive natural numbers. If (n + 1) numbers are randomly selected, then what is the probability that their HCF is 1?

• Q29) If 3 dice are thrown find probability of getting sum as 12?

• Q30) 5 teams play each other once. Probability of each winning is 1/2.what is the probability that no team wins all or loses all?

• OA: 41/44

• Ways in which the first couple can sit together = 2 * 4! (1 couple is considered one unit)
Ways for second couple = 2 * 4!
These cases include an extra case of both couples sitting together
Ways in which both couple are seated together = 2 * 2 * 3! = 4! (2 couples considered as 2 units- so each couple can be arrange between themselves in 2 ways and the 3 units in 3! Ways)
Thus total ways in which at least one couple is seated together = 2 * 4! + 2 * 4! - 4! = 3 * 4!
Total ways to arrange the 5 ppl = 5!
Thus, prob of at least one couple seated together = 3 * 4! / 5! = 3/5
Thus prob of none seated together = 1 - 3/5 = 2/5

• OA: 39/49

• OA: 14/23

• OA : 1/4

• OA : 62.5%

• OA: 1/26

• OA : 3/8

34

57

53

41

50

48

32

60