1.

The probability of choosing at random a number that is divisible by 6 or 8 from among 1 to 90 is equal to   [Pb. CET 2002]

A. \[\frac{1}{6}\]
B. \[\frac{1}{30}\]
C. \[\frac{11}{80}\]
D. \[\frac{23}{90}\]
Answer» E.


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