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What is the probability that an integer chosen randomly from 1000 to 9999 will have 5 in the thousand's place?
A. 1000/9999
B. 1/10
C. 5/10
D. 1/9
E. 5/9
A. 1000/9999
B. 1/10
C. 5/10
D. 1/9
E. 5/9
05 Apr 2020, 04:02
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Evo23
What is the probability that an integer chosen randomly from 1000 to 9999 will have 5 in the thousand's place?
A. 1000/9999
B. 1/10
C. 5/10
D. 1/9
E. 5/9
A. 1000/9999
B. 1/10
C. 5/10
D. 1/9
E. 5/9
total numbers from 1000 to 9999 = 9999-1000 + 1 = 9000
total numbers of type 5 _ _ _ = 10*10*10 = 1000
Required Probability = 1000/9000 = 1/9
Answer : option D
05 Apr 2020, 09:32
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to find the probability that an integer chosen randomly from 1000 to 9999 will have 5 in the thousand's place,
step I : first place is 5
step II : second place can be filled up by 10 ways ( 0,1,2,...,9 digits)
step III : third place can be filled up by 10 ways ( 0,1,2,...,9 digits)
step IV : fourth place can be filled up by 10 ways ( 0,1,2,...,9 digits)
step V : total no of cases : first place can be filled up by 9 ways ( 1,2,...,9 digits)
so, required probability = 10*10*10 / 9*10*10*10 = 1/9
correct answer is D
step I : first place is 5
step II : second place can be filled up by 10 ways ( 0,1,2,...,9 digits)
step III : third place can be filled up by 10 ways ( 0,1,2,...,9 digits)
step IV : fourth place can be filled up by 10 ways ( 0,1,2,...,9 digits)
step V : total no of cases : first place can be filled up by 9 ways ( 1,2,...,9 digits)
so, required probability = 10*10*10 / 9*10*10*10 = 1/9
correct answer is D
