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16 Aug 2016, 06:04
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
76% (01:30) correct
24%
(01:21)
wrong
based on 234
sessions
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Not Attempted Yet
Lottery balls numbered consecutively from one through one hundred are placed in a spinner. If one ball is drawn at random, what is the probability that it will have the number 1 on it exactly once?
A. 9/100
B. 1/10
C. 11/100
D. 19/100
E. 1/5
A. 9/100
B. 1/10
C. 11/100
D. 19/100
E. 1/5
Senthil1981
Joined: 23 Apr 2015
Last visit: 14 Oct 2021
Posts: 225
Given Kudos: 36
Location: United States
Concentration: General Management, International Business
Schools: Booth PT '19 Ross Weekend"19
WE:Engineering (Consulting)
16 Aug 2016, 08:01
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Bunuel
Probability = No of balls with exactly one 1 / total number of balls
= {1,10,12 to 19, 21,31,41,51,61,71,81,91,100}/100
= 19/100
Answer is D
07 Oct 2016, 03:04
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Bunuel
Total possibilities = 100
Break the 1 - 100 into segments
1-9: only one possible outcome = 1
10-19: 1 _, now, in the '_' (units digit) you can have numbers 0 - 9, hence 1*10 = 10. But you need to remove '11' as the questions states '1 exactly once'. 10 - 1 = 9.
20 -29: 2 _, in the '_' only one possible outcome = 1, (remember, in the above case, we have '1' in the tenths place, so we needed to count all possible numbers in the units place)
30 - 39: 3 _, only one possible outcome in units place = 1,
This will repeat for 40-49 to 90-99.
100 will have 1 exactly once.
Ans: \(\frac{(1+9+8+1)}{100} = \frac{19}{100}\)
