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A box contains 90 discs which are numbered from 1 to 90. If one disc is drawn at random from the box, find the probability that it bears (i) a two-digit number, (ii) a number divisible by 5.

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We have, n(S)=90
(i) Let A be the event of getting "a two-digit number".
`therefore` Favourable cases are 10, 11, 12, 13, 14,…, 90.
`therefore" "n(A)=81`
`therefore "P(A)=(n(A))/(n(S))=(81)/(90)=(9)/(10)`
(ii) Let B be the event of getting "a number divisible by 5".
`therefore` Favourable cases are 10, 15, 20, 25, 30, ..., 90.
Let these are n in numbers.
`therefore" "T_(n)=90`
`therefore" "10+(n-1)15=90" "[a_(n) of A.P. = a+ (n-1)d]`
`rArr" "(n-1)5=80 rArr n-1=16 rArr n=17`
`therefore" "n(B)=17`
`therefore" "P(B)=(n(B))/(n(S))=(17)/(90)`

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