gcvbt7t Posted November 15, 2008 Share Posted November 15, 2008 If a number is selected from 10,000 to 99,999, inclusive, what is the probability of getting a number that contain four digits of "5"?(A) 1/1,500 (B) 3/5,000 © 44/100,000 (D) 44/90,000 (E) 52/90,000 Quote Link to comment Share on other sites More sharing options...
e.cartman Posted November 15, 2008 Share Posted November 15, 2008 Imo D. 5555_ The extra digit can be filled by 0to9 except 5 in 9 ways. That can fill in for units, tens, hundreds, thousands places. In 10,000s place, there are only 8 ways. So 9*4+8=44. All possible is 90,000. Quote Link to comment Share on other sites More sharing options...
GMATPAT Posted November 15, 2008 Share Posted November 15, 2008 IMO D AA. (5555X - X can be 1,2,3,4,6,7,8,9) - no. of possible arrangement= 8 * 5!/4! =40 BB.5555X - X can be 0 also . In this case no. of arrangement = 4!/3!=4 Total arrangements with 5555X = 44 Total possible is 90,000 Hence - D OA please.:hmm: Thanks Quote Link to comment Share on other sites More sharing options...
genius_in_the_gene Posted November 16, 2008 Share Posted November 16, 2008 Agree with D Quote Link to comment Share on other sites More sharing options...
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