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07 Mar 2019, 08:09
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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:
75%
(hard)
Question Stats:
62% (02:50) correct
38%
(03:02)
wrong
based on 279
sessions
History
Date
Time
Result
Not Attempted Yet
GMATH practice exercise (Quant Class 18)
Little Julia created a 5-digit integer choosing 5 distinct chips, one by one, among the 7 given ones shown above. "Can you do it in such a way that the three central digits add up to 9?", asked her teacher. And Julia did! If little Max was asked to do the same by his teacher, and Max chooses a correct possibility randomly, what is the probability that both children have chosen exactly the same 5-digit integer?
(A) 1/12
(B) 1/36
(C) 1/54
(D) 1/72
(E) 1/96
Little Julia created a 5-digit integer choosing 5 distinct chips, one by one, among the 7 given ones shown above. "Can you do it in such a way that the three central digits add up to 9?", asked her teacher. And Julia did! If little Max was asked to do the same by his teacher, and Max chooses a correct possibility randomly, what is the probability that both children have chosen exactly the same 5-digit integer?
(A) 1/12
(B) 1/36
(C) 1/54
(D) 1/72
(E) 1/96
eabhgoy
Joined: 12 Apr 2011
Last visit: 14 Jan 2021
Posts: 112
Given Kudos: 85
Location: United Arab Emirates
Concentration: Strategy, Marketing
Schools: CBS '21 Yale '21 INSEAD Jan'21 Intake IMD '21 (A)
GMAT 1: 670 Q50 V31
GMAT 2: 720 Q50 V37
GPA: 3.2
WE:Marketing (Telecommunications)
07 Mar 2019, 11:44
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fskilnik
Lets see the number of 5-digit integers that can be formed such that "three central digits add up to 9".
From the given digits only 5+3+1 = 9, hence the middle three digits are 5,3,1 and the for the remaining two digits we 4 options - 2, 7, 8, 9.
Hence for:
1st digit = 4 ways
5th digit = 3 ways
middle 3 digits can be formed in 3! ways as we have three digits
So total ways = 4*3*3*2*1 = 72 ways
Now if Julia selected a 6 digits number, Max can select the same number with a probability of 1/72 as only way will match out of 72.
Hence D is the correct answer.
General Discussion
07 Mar 2019, 12:45
Kudos
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fskilnik
Very nice, eabhgoy! (Kudos!)
(Just a small correction: "Now if Julia selected a 5 digits number, Max can...")
Here is our official solution:
\(?\,\, = \,\,{1 \over {\# \,\,{\rm{favorable}}\,\,{\rm{sequences}}}}\)
\({?_{temp}}\,\,\, = \,\,\,\# \,\,{\rm{favorable}}\,\,{\rm{sequences}}\)
\(\left\{ \matrix{\\
\,{\rm{3}}\,{\rm{central}}\,{\rm{digits}}\,{\rm{are}}\,\,{\rm{1,3,5}}\,\,\,\, \Rightarrow \,\,\,{{\rm{P}}_{\rm{3}}} = 3!\,\,\,{\rm{possibilities}} \hfill \cr \\
\,{\rm{first}}\,{\rm{and}}\,\,{\rm{last}}\,\,{\rm{digits}}\,\,\,{\rm{:}}\,\,\,{\rm{4}} \cdot {\rm{3}}\,\,{\rm{possibilities}} \hfill \cr} \right.\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,{?_{temp}}\,\,\, = \,\,\,3!\, \cdot 4 \cdot 3\,\, = \,\,\,72\)
The correct answer is (D).
We follow the notations and rationale taught in the GMATH method.
Regards,
Fabio.
