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What is the probability of randomly picking a five-digit number whose

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Bunuel
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What is the probability of randomly picking a five-digit number whose [#permalink] New post   11 Apr 2021, 22:45
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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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Bunuel
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Re: What is the probability of randomly picking a five-digit number whose [#permalink] New post   18 Apr 2021, 22:15
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Bunuel 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.

Answer: B.
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What is the probability of randomly picking a five-digit number whose [#permalink] New post   11 Apr 2021, 23:22
Total 5 digit numbers with different digits= 9*9*8*7*6 { half odd and half even }

Total 5 digit numbers with leftmost digit is 1 and units digit is 9=8*7*6

Answer should be=8*7*6/(9*9*8*7*6/2)

2/81, not matching any option -:) @Bunuel-There seems to be some issue with my approach, please help in identifying.
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Re: What is the probability of randomly picking a five-digit number whose [#permalink] New post   12 Apr 2021, 01:07
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Total number of options for units digit = {1,3,5,7,9} = 5
Total options for first digit = exclude 2 {digit at units place, 0} = 10 - 2 = 8
Total number of possibilities = 8*8*7*6*5

Total favorable options = 1 _ _ _ 9 = 8*7*6

P(E) = 1/(8*5) = 1/40

Ans: B
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Re: What is the probability of randomly picking a five-digit number whose [#permalink] New post   12 Apr 2021, 01:19
Hey, can some one elaborate this more?

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Abi1995
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Re: What is the probability of randomly picking a five-digit number whose [#permalink] New post   12 Apr 2021, 01:21
Lipun wrote:
Total number of options for units digit = {1,3,5,7,9} = 5
Total options for first digit = exclude 2 {digit at units place, 0} = 10 - 2 = 8
Total number of possibilities = 8*8*7*6*5

Total favorable options = 1 _ _ _ 9 = 8*7*6

P(E) = 1/(8*5) = 1/40

Ans: B



What is total no of possibilities here?

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Lipun
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Re: What is the probability of randomly picking a five-digit number whose [#permalink] New post   12 Apr 2021, 01:42
Abi1995 wrote:

What is total no of possibilities here?

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Hi Abi1995,

The total number possibilities here is defined as the set of -
a) five digit,
b) odd numbers,
c) having 5 distinct digits.
all the above constraints are grouped by AND logic.

Using constraint (b), the number of possibilities for units place is reduced to 1,3,5,7,9. Total = 5
So, now the number is : {A} _ _ _ {1,3,5,7,9}
Number of options for A is {1,2,3,4,5,6,7,8,9} - {digit at units place} = 8
The middle 3 digits can be take any values excluding the digit at units place and the digit at A = 8*7*6

So, total possibilities = 8*8*7*6*5

Does it help?

Thanks
Lipun
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Re: What is the probability of randomly picking a five-digit number whose [#permalink] New post   12 Apr 2021, 06:10
There is no restriction saying that digits not be reused so shouldn't the total no of odd digit nos be 8*9*9*9*5?

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Re: What is the probability of randomly picking a five-digit number whose [#permalink] New post   14 Jul 2021, 01:13
I try to solve this problem following this way but can't find out where i made a mistake...
my approach:

The five digit number is =1,3,5,7,9
Question says us that leftmost digit is 1 and unit digit is 9 .

So this two digit is fixed and we have 3 most odd digit from 1 to 9 except 1 and 9 that may rearrange by 3! ways.

So required probability is=3!/5!
=1/20.
where is my mistake? I can't find out.

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Re: What is the probability of randomly picking a five-digit number whose [#permalink] New post   08 Mar 2024, 04:13
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Re: What is the probability of randomly picking a five-digit number whose [#permalink]
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