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# What is the probability of selecting a number that contains

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Manager
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What is the probability of selecting a number that contains [#permalink]

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26 Oct 2005, 10:39
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

What is the probability of selecting a number that contains 5 as one of its digits for the set of integer numbers between { 1, 2, 3, .. 1,000,000}?

Thanks!
Manager
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26 Oct 2005, 15:00
I would have preferred A/C's w/the Q.

Anyway the only pattern I see here is this:

For numbers 1-100:
we have 5, 15, 25, 35, 45, (50-59: ten of these), 65, 75, 85, 95
That's 9*(1) + 10 = 19

1-1000:
This can split into 10 groups of 100 each, and 9 of them behave the same way as the 1-100 shown above. The only exception then is 500-599 (100 of these)
Thats 9*(9*(1) + 10) + 100

Contunuing this pattern..

1-10K:
9*(groups of 1000 see above) + 1K
or 9*(9*(9*(1) + 10) + 100) + 1K

1-100K:
9*(9*(9*(9*(1) + 10) + 100) + 1K) + 10K

1-1000K or 1m:
9*(9*(9*(9*(9*(1) + 10) + 100) + 1K) + 10K) + 100K

So the probability for selecting a number that contains 5 as one of its digits is
[9*(9*(9*(9*(9*(1) + 10) + 100) + 1K) + 10K) + 100K]/1m

I only hope there's an easier way to this
VP
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26 Oct 2005, 16:45
Well I used following method to come up with number of different numbers containing 1 or more 5s.

in 1 digit number: 1 possibility (5)

2 digit number:
x x
let first digit be 5, other can ne any single digit (0-9): 10
1 * 10
the order can be reversed:
1 * 10 * 2
we need to subtract possibility of BOTH numbers being 5 (55, this was included twice in above combinations)

so total number:
1 * 10 * 2 - 1 or
1 * 10 * 2 - 2C2
2 * 10^2 - 2C2
similarly, out of 3 digit numbers, total such numbers are:
1 * 10 * 10 * 3 - (3C2 + 3C3) or
3 * 10^3 - (3C2 + 3C3)

we can got further to 4, 5, 6 digit :
4 digits: 4 * 10^4 - (4C2 + 4C3 + 4C4)
5 digits: 5 * 10^5 - (5C2 + 5C3 + 5C4 + 5C5)
6 digits: 6 * 10^6 - (6C2 + 6C3 + 6C4 + 6C5 + 6C6)

total : 654223

p = 654223 / 1,000,001
Manager
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27 Oct 2005, 04:30
I did the same as mbaqst and got 0'468559

Anyway, the easiest way is (1000000-9^5)/1000000 = 0'468559

the number of numbers without a five is 9^5. The five digits can be anything but 5. Note that in this calculation we are including "00000" but as we are working with five digit numbers, we are failing to consider 1000000, therefore, we should add one (1000000) and subtract one (00000) to agree with the stem [1,1000000].
VP
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27 Oct 2005, 07:51
jdtomatito wrote:
I did the same as mbaqst and got 0'468559

Anyway, the easiest way is (1000000-9^5)/1000000 = 0'468559

the number of numbers without a five is 9^5. The five digits can be anything but 5. Note that in this calculation we are including "00000" but as we are working with five digit numbers, we are failing to consider 1000000, therefore, we should add one (1000000) and subtract one (00000) to agree with the stem [1,1000000].

can you explain bod portion please.
Are you considering ONLY 5 digit numbers?
(a) all the numbers that include 5 can be any digit 1 through 6 IMO eg. 5, 15, 35, 55, 555,525,....,555555,555557

(b) we need to go upto 6 digits (not 5)
Manager
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27 Oct 2005, 17:02
Sorry guys, this one is from a probability text book, that's why I couldn't give multiple choice answers. I only have the OA from the back of the book, which is:

0.469
Manager
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28 Oct 2005, 00:29
My mistake, I meant 9^6, anyway, the result is correct.

Ok, I'll try to explain,

We need to consider all numbers between 1 and 1000000. We know that 1000000 does not contain a 5 and that 0 does not contain a five either.
Therefore, we can safely consider all numbers between 0 and 999999 instead.

We can calculate the probability to pick one number with at least a five in it as (1 - the probability to pick a number with no fives in it).

First digit can be 012346789= 9 possibilities
Second digit can be 012346789 = 9 possibilities
And so on...

Numbers with no fives 9*9*9*9*9*9 = 9^6

Therefore P(pick a number with at least a five) = 1-(9^6/1000000)

= 0'468559 aprox. 0.469 that is the official answer.

I hope this is clearer
VP
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28 Oct 2005, 08:44
jdtomatito wrote:
My mistake, I meant 9^6, anyway, the result is correct.

Ok, I'll try to explain,

We need to consider all numbers between 1 and 1000000. We know that 1000000 does not contain a 5 and that 0 does not contain a five either.
Therefore, we can safely consider all numbers between 0 and 999999 instead.

We can calculate the probability to pick one number with at least a five in it as (1 - the probability to pick a number with no fives in it).

First digit can be 012346789= 9 possibilities
Second digit can be 012346789 = 9 possibilities
And so on...

Numbers with no fives 9*9*9*9*9*9 = 9^6

Therefore P(pick a number with at least a five) = 1-(9^6/1000000)

= 0'468559 aprox. 0.469 that is the official answer.

I hope this is clearer

Very nice explanation. Thanks.
Manager
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28 Oct 2005, 09:43
Excellent jdtomatito! Easy way of solving the Q and great explanation..
28 Oct 2005, 09:43
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