Prob Question

There are 9 seats in a row in a classroom. 5 students go into the classroom in which one of them is the monitor and one of them is the monitress.

(a) Find the probability that both the monitor and the monitress do not sit at the two ends.

The required probability
= 1 - Pr(both the monitor and the monitress sit at the two ends)
= 1 - ( 2 × ₉₋₂P₅₋₂ / ₉P₅ )
= 1 - ( 2 × ₇P₃ / ₉P₅ )
= 1 - ( 2 × 7×6×5 ) / ( 9×8×7×6×5 )
= 35/36

Here, 2 means the possible cases of both the monitor and the monitress sitting at the end.
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₇P₃ means the number arrangements of the remaining 3 students in the remaining 7 seats.

₉P₅ means the total number of arrangements of 5 students in the 9 seats.


(b) Find the probability that the monitor and the monitress sit next to each other.

The required probability
= 8 × 2 × ₉₋₂P₅₋₂ / ₉P₅
= 8 × 2 × ₉₋₂P₅₋₂ / ₉P₅
= 8 × 2 × ₇P₃ / ₉P₅
= 8 × 2 × 7×6×5 / ( 9×8×7×6×5 )
= 2/9

Here, 8 means the possible cases of both the monitor and the monitress sitting next to each other.
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2 means the swapping between the monitor and the monitress.

₇P₃ means the number arrangements of the remaining 3 students in the remaining 7 seats.

₉P₅ means the total number of arrangements of 5 students in the 9 seats.

2014-03-15 22:26:43 補充：
不好意思，(a) 部分可能我誤會了題目。

以上的作答是： not (both the monitor and the monitress sit at the two ends)

你的問題應是 both the monitor and the monitress do not sit at the two ends.

所以應該解答如下：

　Pr(both the monitor and the monitress do not sit at the two ends)
= ₇P₂ × ₉₋₂P₅₋₂ / ₉P₅
= ₇P₂ × ₇P₃ / ₉P₅

2014-03-15 22:29:00 補充：
= (7×6) × (7×6×5) / ( 9×8×7×6×5 )
= 7/12

Reasons:
(1) First choose two seats in the middle (7 seats) for the monitor and the monitress.
(2) Assign the remaining 3 students in the remaining 7 seats.

2014-03-16 16:32:56 補充：
很感謝 andrew 提供寶貴的意見！

Thanks!

☆ヾ(◕‿◕)ノ

M/m : the monitor and/or the monitress
5 students take seats without restriction = P(9, 5)

(a)
Method 1 :
- both M/m take seats from middle 7 seats, P(7,2)
- The rest 3 students take 3 of the rest 7 seats, P(7,3)

The req'd probability
= P(7,2) x P(7,3) / P(9,5)
= (7!/5!) x (7!/4!) / (9!/4!)
= 7/12

2014-03-16 09:37:55 補充：
(a)
Method 2 :

Case 1 :
- Both M/m take seats at two ends, P(2,2)
- The rest 3 students take 3 of the rest 7 seats, P(7,3)

Case 2 :
- Either M/m takes seat at 2 ends, C(2,1) x P(2,1)
- The rest M/m takes 1 of the rest 7 middle seats, P(7,1)
- The rest 3 students take 3 of the rest 7 seats, P(7,3)

2014-03-16 09:38:16 補充：
The required probability
= 1 - P(case 1) - P(case 2)
= 1 - [P(2,2) x P(7,3) / P(9,5)] - [C(2,1) x P(2,1) x P(7,1) x P(7,3) / P(9, 5)]
= 1 - (2 x 210 / 15120) - (2 x 2 x 7 x 210 / 15120)
= 1 - (1/36) - (14/36)
= 7/12

2014-03-16 09:38:39 補充：
(b)
- Buddle the pair M/m, internal permutation = P(2,2)
- 4 units take 8 seats, P(8,4)

The required probability
= P(2,2) x P(8,4) / P(9,5)
= 2 x 1680 / 15120
= 2/9