Numbers 12, 13, 14, ..., 101, 102 and 103 are written on slips (one number on one slip) and put in a box. A slip is then taken out from the box at random. What is the probability that the selected slip bears a number divisible by 7?
\(\frac{13}{92}\)
The numbers on the slips run from 12 to 103, so the total number of slips is \(103-12+1=92\).
The multiples of 7 in this range start at 14 and end at 98, since \(7\times2=14\) and \(7\times14=98\), while \(7\times15=105\) exceeds 103.
The number of such multiples is \(\frac{98-14}{7}+1=13\).
So the probability of drawing a slip divisible by 7 is \(\frac{13}{92}\).
Hence, the required probability is \(\frac{13}{92}\).
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