A) \[\frac{1}{6}\]
B) \[\frac{1}{90}\]
C) \[\frac{1}{30}\]
D) \[\frac{23}{90}\]
Correct Answer: D
Solution :
Total numbers \[=90\] Numbers divisible by \[6=6,\,12,18,24,30,36.\] \[42,48,54,60,66,72,78,84,90\] Number divisible by \[8=8,16,24,32,40,48,\] \[56,64,72,80,88\] Total number of numbers divisible by 6 or 8 \[=15+11-3=23\] \[\therefore \] Required probability \[=\frac{23}{90}\]
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