<p>An integer from 100 through 999, inclusive, is to bechosen at random. What is the probability that the number chosen will have 0 as at least 1 digit?</p>
<p>(a) 19/900
(b)81/900
(c) 90/900
(d) 171/900
(e) 271/1000</p>
<p>After looking at this problem for a minute I thought the answer was (C), but it was not correct. Does anybody have an easy way to solve probability problems like that? Instructions that I was coming across so far keep confusing me..</p>
<p>well, 100 to 110, 201 to 209, 301 to 309, 401 to 409, 501 to 509, 601 to 609, 701 to 709, 801 to 809 and 901 to 909 all have a zero. Equaling 11+(8<em>8) digits.
every tenth number from 120 to 990 has one zero, add those up and then add the sum to 11+(8</em>8) and that should be your answer</p>
<p>the answer id d). From 100 to 110 - 11. 120 130 140 150 160 170 180 190 - 8. so 11 no. between 100 to 199. mulyiply 11*9 = 171. 9 because 110-199, 200-299 etc…</p>
<p>Dd.</p>
<p>999-100+1=900 numbers in total (denominator)</p>
<p>9(100, 200, 300…900) + 81([10, 20, 30…90] x 9) + ([01, 02, 03…09] x 9) = 171 numbers with a zero in it (numerator)</p>
<p>So the answer is 171/900, or Dd.</p>