ProbabilityHard
Question
Three numbers are chosen at random without replacement from {1, 2, 3,......, 10}. The probability that the minimum of the chosen numbers is 3 or their maximum is 7 is:
Options
A.11/40
B.9/40
C.
D.none
Solution
n(S) = ways of selecting 3 number from 10 is 10C3
n(E) → n (A ∪ B) where A → min. number chosen is 3
n(A) = 7C2
B → max number chosen is 7
n(B) = 6C2 also n(A ∩ B) = 3C1 = 3
n(E) = 7C2 + 6C2 - 3
n(E) → n (A ∪ B) where A → min. number chosen is 3
n(A) = 7C2
B → max number chosen is 7
n(B) = 6C2 also n(A ∩ B) = 3C1 = 3
n(E) = 7C2 + 6C2 - 3
Create a free account to view solution
View Solution FreeMore Probability Questions
A bag contains 5 black and 3 blue balls. Balls are drawn (without replacement) one by one. The probability of getting bl...Three identical dice are rolled. The probbability that the same number will appear on each of t hem, is...A bag contains 10 balls out of which k are red and $(10 - k)$ are black, where $0 \leq k \leq 10$. If three balls are dr...If a, b and c are three numbers (not necessarily different) chosen randomly and with replacement from the set {1, 2, 3, ...A coin is tossed indefinitely. Probability that the fourth head occurs before the sixth toss is equal to -...