SOLUTION: Hi, I'm trying to solve this problem: If a 3-digit number (000 to 999) is chosen at random, find the probability that exactly one digit will be less than 2. I got to the point

Question 15704: Hi,
I'm trying to solve this problem: If a 3-digit number (000 to 999) is chosen at random, find the probability that exactly one digit will be less than 2.
I got to the point of (1/10)^2 * 2/10 but I'm not sure or am I answering another problem. Please help asap.
Found 2 solutions by venugopalramana, biplavmehta:
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
If a 3-digit number (000 to 999) is chosen at random, find the probability that exactly one digit will be less than 2.
there are some things not clear ..a 3 digit number if you say ,it can be only 100 to 999 ,not 000 to 999..please check how/where you got the idea. any way i will give the method
total 3 digit numbers = 999-100+1 = 900
now less than 2 we can have 1 or 0 . but in the first digit we can not have zero.hence numbers with 1 in the first digit can be
1(1st.digit)*8(second digit)*8(3rd.digit)= 64..(second & 3rd. digits will have to be between 2 and 9 ..that is 8 possibilities)
Similarly with 1 or zero in the second digit we can have
8*2*8 = 128
Similarly with 1 or zero in the 3rd. digit we can have
8*8*2 = 128
so total possibilities =64+128+128=320
probability =320/900 =32/90 =16/45

Answer by biplavmehta(3) (Show Source): You can put this solution on YOUR website!
let 1st digit is less then 2 and remaining more then equal to 2. then we have 2*8*8 favorable no. of cases.
similarly for 2nd and 3rd digit.
so the total no. of favorable cases = 2*8*8 + 8*2*8 + 8*8*2 = 384.
total no. of cases= 10*10*10=1000
ans= 384/1000.
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