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05 Jan 2021, 08:39
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
80% (01:01) correct
20%
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wrong
based on 143
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Eleven chairs are numbered 1 through 11. Four girls and seven boys sit on these chairs at random. What is the probability that chair 5 is occupied by a boy?
A. 1/22
B. 1/11
C. 4/11
D. 5/11
E. 7/11
A. 1/22
B. 1/11
C. 4/11
D. 5/11
E. 7/11
rocky620
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05 Jan 2021, 08:56
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As there are 7 boys, each Chair has a probability of 7/11 to be occupied by a boy.
IMO Option E
IMO Option E
05 Jan 2021, 09:16
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Based on the concept of “Symmetry” in the Arrangements:
If we put Girl 1 in seat 5, and shuffled the other 10 unique ppl around in every way possible, the no of Arrangements would equal
=
The no of arrangements when we put Girl 2 in seat 5, and shuffle the other 10 unique ppl around in every way possible. These no of Arrangements will equal
=
The no of Arrangements if we put Boy 1 in seat 5, and shuffle the other 10 ppl around in every way possible
.......And so on for the 11 people.
Thus, out of the 11! Distinct Arrangements, we will have 7 equal “countings” of Arrangements that have a boy in the 5th seat as well as 4 equal “countings” of Arrangements that have a girl in the 5th seat.
These 11 scenarios will all have the same no of Arrangements.
Probability that a Boy occupies the 5th Seat =
No of favorable outcomes = (7/11) * 11! ——-> which is 7/11th of all the unique arrangements possible
_____________________________
Divided by
_____________________________
Total Possible Outcomes = 11! Distinct Arrangements
= [ (7/11) * (11!) ] / (11!)
= 7/11
E
Posted from my mobile device
If we put Girl 1 in seat 5, and shuffled the other 10 unique ppl around in every way possible, the no of Arrangements would equal
=
The no of arrangements when we put Girl 2 in seat 5, and shuffle the other 10 unique ppl around in every way possible. These no of Arrangements will equal
=
The no of Arrangements if we put Boy 1 in seat 5, and shuffle the other 10 ppl around in every way possible
.......And so on for the 11 people.
Thus, out of the 11! Distinct Arrangements, we will have 7 equal “countings” of Arrangements that have a boy in the 5th seat as well as 4 equal “countings” of Arrangements that have a girl in the 5th seat.
These 11 scenarios will all have the same no of Arrangements.
Probability that a Boy occupies the 5th Seat =
No of favorable outcomes = (7/11) * 11! ——-> which is 7/11th of all the unique arrangements possible
_____________________________
Divided by
_____________________________
Total Possible Outcomes = 11! Distinct Arrangements
= [ (7/11) * (11!) ] / (11!)
= 7/11
E
Posted from my mobile device
