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05 Jan 2005, 20:01
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In a lottery there are 900 tickets numbered consecutively 100, 101, ..., 998, 999. If a ticket is drawn at random, what is the probability that
(i) the three digits are neither in ascending or descending consecutive order (e.g. not of the form 345 or 765),
(ii) the three digits read the same backwards and forwards?
Please provide your calculation with your answer.
(i) the three digits are neither in ascending or descending consecutive order (e.g. not of the form 345 or 765),
(ii) the three digits read the same backwards and forwards?
Please provide your calculation with your answer.
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05 Jan 2005, 20:50
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i) 1-p(the 3 digits are in ascending or descending order)
there are 7 such numbers in ascending order and 7 such numbers in descending vis a vis
123, 234, 345, 456, 567, 678 & 789 (the same set of numbers in reverse order)
p = 1 - 14/900 = 886/900
ii) there are 9 such numbers
111, 222 ..... 999
p = 9/900 = 1/100
05 Jan 2005, 21:19
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A) 886/900
Nos which are ascending or descending
100s = 1 i.e. 123
200s = 1 i.e. 234
300s = 2 i.e. 345 and 321
400s = 2
500s = 2
600s = 2
700s = 2
800s = 1 only 876
900s = 1 only 987
So P(neither descending or ascending) = 1-14/900 = 886 / 900
B) 1/10
100s = 10 i.e. 101, 111, 121, 131, 141, 151, 161, 171, 181, 191
200s = 10 i.e. 202, 212, 222, 232, 242, 252, 262, 272, 282, 292
300s = 10 i.e. 303, 313 ...
400s = 10
500s = 10
600s = 10
700s = 10
800s = 10
900s = 10
p(e) = 90/900 = 1/10
Nos which are ascending or descending
100s = 1 i.e. 123
200s = 1 i.e. 234
300s = 2 i.e. 345 and 321
400s = 2
500s = 2
600s = 2
700s = 2
800s = 1 only 876
900s = 1 only 987
So P(neither descending or ascending) = 1-14/900 = 886 / 900
B) 1/10
100s = 10 i.e. 101, 111, 121, 131, 141, 151, 161, 171, 181, 191
200s = 10 i.e. 202, 212, 222, 232, 242, 252, 262, 272, 282, 292
300s = 10 i.e. 303, 313 ...
400s = 10
500s = 10
600s = 10
700s = 10
800s = 10
900s = 10
p(e) = 90/900 = 1/10
