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
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