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# What is the probability of randomly picking a five-digit

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Senior Manager
Joined: 21 Oct 2013
Posts: 414
What is the probability of randomly picking a five-digit  [#permalink]

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27 Jul 2014, 10:05
2
12
00:00

Difficulty:

75% (hard)

Question Stats:

60% (02:57) correct 40% (01:57) wrong based on 179 sessions

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What is the probability of randomly picking a five-digit number whose leftmost digit is 1 and units digit is 9 from all the odd five-digit numbers that have different digits?

A. 1/80
B. 1/40
C. 1/25
D. 1/20
E. 1/10
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Joined: 02 Sep 2009
Posts: 56304
Re: What is the probability of randomly picking a five-digit  [#permalink]

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27 Jul 2014, 15:23
7
2
goodyear2013 wrote:
What is the probability of randomly picking a five-digit number whose leftmost digit is 1 and units digit is 9 from all the odd five-digit numbers that have different digits?

A. 1/80
B. 1/40
C. 1/25
D. 1/20
E. 1/10

The number of all the odd five-digit numbers that have different digits is 5*8*8*7*6: 5 odd digits for the 5th (last) digit, 8 options for the first digit (no 0 and no the odd digit we used for the last digit), 8 options for the second digit (no two digits we already used), 7 options for the third digit (no three digits we already used), and 6 options for the fourth digit (no four digits we already used). We are starting from the digit for which we have most constraints, then the next digit with constraints and so on.

The number of five-digit numbers whose leftmost digit is 1 and units digit is 9 (1XXX9) that have different digits is 8*7*6.

P = (favorable)/(total) = (8*7*6)/(5*8*8*7*6) = 1/40.

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Joined: 03 Oct 2012
Posts: 26
Concentration: Strategy, General Management
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Re: What is the probability of randomly picking a five-digit  [#permalink]

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27 Jul 2014, 19:58
1
Great explanation Bunuel (as usual)

Is the tag correct though? I didn't think this was a sub-600 question at all!
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Re: What is the probability of randomly picking a five-digit  [#permalink]

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27 Jul 2014, 23:49
SachinWordsmith wrote:
Great explanation Bunuel (as usual)

Is the tag correct though? I didn't think this was a sub-600 question at all!

Edited the tag. Thank you.
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What is the probability of randomly picking a five-digit  [#permalink]

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09 Oct 2015, 09:03
What a silly mistake I did. In numerator I kept the repeated digits since it isn't mentioned explicitly albeit obvious that the number we pick would also have different digits. I kept it as 10*10*10/8*8*7*6*5
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What is the probability of randomly picking a five-digit  [#permalink]

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13 Jan 2016, 19:41
2
oh damn..I did a crucial mistake to consider that all digits must be odd got to 1/20 (

only after I clicked the answer..I re-read the question and understood where I made the error
we have 10 digits, last one must be odd, thus 5 possible ways. first one has to be any digit except for 0 and last one, thus we have 8 ways to pick first digit. then second, we can pick from 8 ways (first one is picked, and last one too, but we did not consider 0 in the first two cases, hence we have here 8 ways), then 7 for the third digit, and 6 for the fourth.
we have 8*8*7*6*5

for a number to have 1 as first digit, and 9 as the last, no repeating digits: we have 1*8*7*6*1

THEN, the probability is
8*7*6/8*8*7*6*5 = 1/40...
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Re: What is the probability of randomly picking a five-digit  [#permalink]

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08 May 2017, 14:38
This was a very tricky question. I got tricked and misunderstood that all 5 digits were meant to be odd. So I ended up with 1/20. whoops. Great explanation bunnel
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Re: What is the probability of randomly picking a five-digit  [#permalink]

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17 May 2017, 00:20
Bunuel wrote:
goodyear2013 wrote:
What is the probability of randomly picking a five-digit number whose leftmost digit is 1 and units digit is 9 from all the odd five-digit numbers that have different digits?

A. 1/80
B. 1/40
C. 1/25
D. 1/20
E. 1/10

The number of all the odd five-digit numbers that have different digits is 5*8*8*7*6: 5 odd digits for the 5th (last) digit, 8 options for the first digit (no 0 and no the odd digit we used for the last digit), 8 options for the second digit (no two digits we already used), 7 options for the third digit (no three digits we already used), and 6 options for the fourth digit (no four digits we already used). We are starting from the digit for which we have most constraints, then the next digit with constraints and so on.

The number of five-digit numbers whose leftmost digit is 1 and units digit is 9 (1XXX9) that have different digits is 8*7*6.

P = (favorable)/(total) = (8*7*6)/(5*8*8*7*6) = 1/40.

Attached is a graphic illustration of what I believe is Bunuel's explanation
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IMG_0521.JPG [ 748.17 KiB | Viewed 3091 times ]

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Re: What is the probability of randomly picking a five-digit  [#permalink]

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09 Aug 2018, 05:35
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Re: What is the probability of randomly picking a five-digit   [#permalink] 09 Aug 2018, 05:35
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# What is the probability of randomly picking a five-digit

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