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11 Apr 2021, 22:45
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B
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Difficulty:
75%
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Question Stats:
56% (02:30) correct
44%
(02:15)
wrong
based on 127
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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
A. 1/80
B. 1/40
C. 1/25
D. 1/20
E. 1/10
18 Apr 2021, 22:15
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Bunuel
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
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.
General Discussion
11 Apr 2021, 23:22
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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.
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.
