## Karl Proposes a Probability Question. Risky!

Posted by Karl Sharman on January 8, 2011 – 12:10 pm

If we consider 2-digit numbers, there are exactly 18 numbers that contain the digit one: 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 21, 31, 41, 51, 61, 71, 81, 91

As there are ninety 2-digit numbers, the probability of a 2-digit number containing a one is 18/90 = 1/5. If I have got this wrong, then the answer I have is wrong, and some one better tell me fairly quickly so I can remove the problem!

What proportion of 3-digit numbers contain the digit one?

January 8th, 2011 at 1:04 pm

seems to be 28%

900 3 didit nos. 100-999

all 100-199 =100

the 18 mentioned above for each 200-900 =144

and also 201, 301…..901 =8

total 252/900 = 28%

January 8th, 2011 at 1:07 pm

There are 900 three digit numbers. The first one-hundred have a one in them, for the other sets of one-hundred, each has 19 of that set containing a 1 digit. 8*19 is 152, add 100 and divide by 900, e.g. 252/900 is 28%

January 8th, 2011 at 3:56 pm

Hi Karl, you got your question right.

The rest is as Nathan said.

So what is the probabilty that a number between 100 and 999 (inclusive) contains two 1’s, given that it contains at least one 1?

January 8th, 2011 at 8:16 pm

Of the 900 numbers, 252 contain at least one one as shown above. of those, 27 have at least two more more one’s. This give a 3/28 or 10.7% chance that a number has at least two ones, given that you know it contains at least one one.

January 8th, 2011 at 10:38 pm

13/45 or 260 in 900

January 9th, 2011 at 10:09 pm

I got 280 of 900 or 14/45

January 10th, 2011 at 6:42 am

Chris – Post 3 – I think that there is a probability of…

252 with 1 number 1

27 with 2 number 1’s

For any 3 digit number there are 252 possible and equally likely combinations where at least 1 number is a 1.

Only 27 of those combinations has 2 number ones in it.

So…. deep breath…. 1 in 9.333?

January 10th, 2011 at 9:30 am

The answer is 131/450

January 11th, 2011 at 1:07 am

well there are 900 3 digit numbers and there are.. 1 10 10 + 8 1 1 + 8 10 1 + 8 1 10 + 1 = 269 with the number one in at least one spot. I am 9% sure that that is right.. i am in math 12 and we are working on probabilities right now. im going to ask the teacher tomorrow though

January 11th, 2011 at 1:14 am

oh i am wrong. i realized that i double counted a few things.. i get 252 now

January 14th, 2011 at 7:48 am

Hi Karl,

Could you update the “Adds Up To a Thousand” puzzle with a solution please?

Nathan, Wiz and I all have different answers for the number of unique triplet combinations and it’s been a few days with no new posts (I think Chris’s “Girls Girls Girls” post has had all the attention lately).

Cheers.