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3 digit probability

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by neoreaves » Mon Apr 05, 2010 10:02 am
If each of the digits can be used, as many times as necessary, what is the probability of creating a three-digit number with three consecutive even digits in order from left to right?
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by thephoenix » Mon Apr 05, 2010 10:15 am
even digit=0,2,4,6,8
3 consecutive even digits are 1st--->0,2,4 ; 2nd---->2,4,6;3rd---->4,6,8;
for 1st set tot # can be formed in 2*3*3=18 ways
for 2nd set tot # can be formed in 3^3=27 ways
for 3rd set tot # can be formed in 3^3=27 ways
tot=18+27+27=72
wats the ans
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by eaakbari » Mon Apr 05, 2010 11:35 am
A little confused again with the wording. Does it mean the number has to be in consecutive order or random.

If random answer is 2/25 or if in order 2/900
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by pops » Mon Apr 05, 2010 8:56 pm
neoreaves wrote:If each of the digits can be used, as many times as necessary, what is the probability of creating a three-digit number with three consecutive even digits in order from left to right?
Number of three digit numbers = 9*10*10=900
Number of three digit numbers with three consecutive even digits from left to right=246,468 (note that 024 is not 3 digit number and 680 does not mean 3 consecutive even digits)
hence 2/900 = 1/450
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