A) zero
B) \[\frac{55}{1000}\]
C) \[\frac{33}{1000}\]
D) \[\frac{44}{1000}\]
Correct Answer: B
Solution :
Since, in 1000 pages there are 55 pages in which the sum of digits of page number is 9. \[\therefore \] Required probability\[=\frac{55}{1000}\].You need to login to perform this action.
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