Question 7E-9 Three friends and seven other people are randomly seated in a row. Specify an appropriate sample space to answer the following two questions.
(a) What is the probability that the three friends will sit next to each other?
(b) What is the probability that exactly two of the three friends site next to each other?
Solution to 7E-9: Take as sample space the set of all possible orderings of the ten people on the ten seats. The sample space has 10! equally likely outcomes.
(a) The number of ways to choose three adjacent seats from the ten seats is 8. Hence the number of outcomes for which the three friends are seated next to each other is:
. . . . .\(\displaystyle 8\, \times\, 3!\, \times\, 7!\)
and so the probability that the three friends will sit next to each other is:
. . . . .\(\displaystyle \dfrac{8\, \times\, 3!\, \times\, 7!}{10!}\, =\, \dfrac{1}{15}\)
(b) The number of ways to choose two friends from the three friends is:
. . . . .\(\displaystyle \large{ \binom{3}{2} }\)
The number of outcomes for which the two chosen friends sit next to each other at one of the ends of the row and the third friend has a non-adjacent seat is:
. . . . .\(\displaystyle 2\, \times\, 7\, \times\, 2!\, \times\, 7!\)
while the number of outcomes for which the two chosen friends site next to each other but not at one of the ends of the row and the third friends has a non-adjacent seat is:
. . . . .\(\displaystyle 7\, \times\, 6\, \times\, 2!\, \times\, 7!\)
Hence the probability that exactly two of the three friends will sit next to each other is:
. . . . .\(\displaystyle \large{ \dfrac{\binom{3}{2}\, \left[\, 2\, \times\,7 \,\times\, 2! \,\times\, 7!\, +\, 7 \,\times\, 6 \,\times\, 2! \,\times\, 7!\, \right]}{10!}\, =\, \dfrac{7}{15} }\)
I don't understand how the answer to (b) was reached. I don't understand where any of the numbers come from in defining how many possibilities there are for each stated outcome. Could someone elaborate into how this was done?
Thanks.
(a) What is the probability that the three friends will sit next to each other?
(b) What is the probability that exactly two of the three friends site next to each other?
Solution to 7E-9: Take as sample space the set of all possible orderings of the ten people on the ten seats. The sample space has 10! equally likely outcomes.
(a) The number of ways to choose three adjacent seats from the ten seats is 8. Hence the number of outcomes for which the three friends are seated next to each other is:
. . . . .\(\displaystyle 8\, \times\, 3!\, \times\, 7!\)
and so the probability that the three friends will sit next to each other is:
. . . . .\(\displaystyle \dfrac{8\, \times\, 3!\, \times\, 7!}{10!}\, =\, \dfrac{1}{15}\)
(b) The number of ways to choose two friends from the three friends is:
. . . . .\(\displaystyle \large{ \binom{3}{2} }\)
The number of outcomes for which the two chosen friends sit next to each other at one of the ends of the row and the third friend has a non-adjacent seat is:
. . . . .\(\displaystyle 2\, \times\, 7\, \times\, 2!\, \times\, 7!\)
while the number of outcomes for which the two chosen friends site next to each other but not at one of the ends of the row and the third friends has a non-adjacent seat is:
. . . . .\(\displaystyle 7\, \times\, 6\, \times\, 2!\, \times\, 7!\)
Hence the probability that exactly two of the three friends will sit next to each other is:
. . . . .\(\displaystyle \large{ \dfrac{\binom{3}{2}\, \left[\, 2\, \times\,7 \,\times\, 2! \,\times\, 7!\, +\, 7 \,\times\, 6 \,\times\, 2! \,\times\, 7!\, \right]}{10!}\, =\, \dfrac{7}{15} }\)
I don't understand how the answer to (b) was reached. I don't understand where any of the numbers come from in defining how many possibilities there are for each stated outcome. Could someone elaborate into how this was done?
Thanks.
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