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13 Jul 2015, 23:09
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Q1.
A group of 5 students bought movie tickets in one row next to each other. If Bob and Lisa are in this group, what is the probability that Bob and Lisa will each sit next to only one of the four other students from the group?
(A) 5%
(B) 10%
(C) 15%
(D) 20%
(E) 25%
Q2.A group of 5 students bought movie tickets in one row next to each other. If Bob and Lisa are in this group,
what is the number of ways of seating if both of them will sit next to only one other student from the group?
Hi Bunnel ,
Is there any difference between the two question on calculating the number cases in Bob and Lisa condition mentioned in the above question.
Please need your input .
Thanks .
A group of 5 students bought movie tickets in one row next to each other. If Bob and Lisa are in this group, what is the probability that Bob and Lisa will each sit next to only one of the four other students from the group?
(A) 5%
(B) 10%
(C) 15%
(D) 20%
(E) 25%
Q2.A group of 5 students bought movie tickets in one row next to each other. If Bob and Lisa are in this group,
what is the number of ways of seating if both of them will sit next to only one other student from the group?
Hi Bunnel ,
Is there any difference between the two question on calculating the number cases in Bob and Lisa condition mentioned in the above question.
Please need your input .
Thanks .
Archived Topic
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This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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14 Jul 2015, 00:10
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abhisheknandy08
Solution:
If bob and Lisa have to sit next to only of the other 4 members of the group then Bob and Lisa must sit at the extremes
Bob and Lisa can sit at the extreme two positions in 2! ways
Remaining 3 persons can sit between bob and lisa in 3! ways
i.e. Total Favorable Seating arrangements = 2!*3! = 12
Total Possible Seating arrangements of 5 people = 5! = 120
Probability = 12/120 = 1/10 = 10%
Answer: Option B
abhisheknandy08
This language is confusing and lacks clarity however my interpretation worked like this
This Question means that Both must sit next to only one Common Student i.e. both Bob and Lisa are separated by just one seat between them
So consider the Pair of Bob and Lisa separated by one seat between them and this pair can move in 3! ways and bob and Lisa can exchange seats in 2! ways
i.e. Ways in which Bob and Lisa can sit = 3*2! = 6 ways
Remaining 3 persons can sit in 3! ways
i.e. Total Favorable Seating arrangements = 6*3! = 36
and Total Possible Seating arrangements of 5 people = 5! = 120
Probability = 36/120 = 3/10 = 30%
14 Jul 2015, 00:35
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abhisheknandy08
PLEASE READ CAREFULLY AND FOLLOW: rules-for-posting-please-read-this-before-posting-133935.html
Q#1: a-group-of-5-students-bought-movie-tickets-in-one-row-next-47450.html
Q#2: a-group-of-5-students-bought-movie-tickets-in-one-row-next-t-162572.html
TOPIC LOCKED.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
