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 2digit, 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