

MCQOPTIONS
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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. | |