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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
41% (01:50) correct
59%
(01:57)
wrong
based on 331
sessions
History
Date
Time
Result
Not Attempted Yet
A box contains 100 tickets marked 1 to 100. One ticket is picked at random. What is the probability that the number on the ticket will be divisible by either 2 or 3, but not 6?
A. 83/100
B. 67/100
C. 51/100
D. 49/100
E. 33/100
The official answer is:
A. 83/100
B. 67/100
C. 51/100
D. 49/100
E. 33/100
The official answer is:
51/100[
The official solution is the following:
All possibilities = 100
Numbers divisible by 2 = 50 ... A
Numbers divisible by 3 = 33 ... B
Numbers divisible by 6 = 16 ... C
Positive outcomes = A + B - 2*C
Probability = 51/100
Can someone please explain why they subtract C twice?
Thank you.
The official solution is the following:
All possibilities = 100
Numbers divisible by 2 = 50 ... A
Numbers divisible by 3 = 33 ... B
Numbers divisible by 6 = 16 ... C
Positive outcomes = A + B - 2*C
Probability = 51/100
Can someone please explain why they subtract C twice?
Thank you.
07 Feb 2011, 03:24
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Bookmarks
nonameee
A box contains 100 tickets marked 1 to 100. One ticket is picked at random. What is the probability that the number on the ticket will be divisible by either 2 or 3, but not 6?
Total = 100 numbers.
Numbers divisible by 2 = 50 ... A
Numbers divisible by 3 = 33 ... B
Numbers divisible by 6 = 16 ... C
Now, both A (numbers divisible by 2) and B (numbers divisible by 3) contain C (numbers divisible by 6) so to get the numbers which are divisible by either 2 or 3, but not 6 we should subtract C twice, once for A and once for B: A + B - 2*C= 51
Probability = 51/100
If it were: "What is the probability that the number on the ticket will be divisible by either 2 or 3 (or both)?" Then we would subtract C only once to get rid of double counting.
07 Feb 2011, 06:09
Kudos
Bookmarks
I thought it to be 67/100 and realized my mistake.
Count of numbers divisible by 2 = 50. In these 50 numbers there are 16 numbers that are divisible by 6 and we must take them out. 50-16=34
Count of numbers divisible by 3 = 33. In these 33 numbers there are 16 numbers that are divisible by 6 and we must take them out. 33-16=17
Total numbers divisible by 2 or 3 but not 6 = 34+17=51
Probability = 51/100.
Good one.
Count of numbers divisible by 2 = 50. In these 50 numbers there are 16 numbers that are divisible by 6 and we must take them out. 50-16=34
Count of numbers divisible by 3 = 33. In these 33 numbers there are 16 numbers that are divisible by 6 and we must take them out. 33-16=17
Total numbers divisible by 2 or 3 but not 6 = 34+17=51
Probability = 51/100.
Good one.
