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

Question

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

Bunuel · Accepted Answer

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.

sumitkrocks · Answer

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.

Lipun · Answer

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

Abi1995 · Answer

What is total no of possibilities here?

Posted from my mobile device

Lipun · Answer

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

finisher009 · Answer

reasoning "half odd and half even," is  not valid  when digits are unique and restrictions exist.

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

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

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