F6 Probability Q12

Method1 :
No. of ways if 5 students sit randomly
= P(10,5)
= 10!/5!
= 10 x 9 x 8!/5!
= 90 x 8!/5!

(Consider the ways that Amy and Billy sitting next to each other.)
(The 10 seats : XYYYYYYYYX)
(If Amy sits on one of the 2 X, Billy should take the only seat next to Amy,and the other three take 3 of the rest 8 seats.)
(No. of ways in the above situation = P(2,1) x P(1, 1) x P(8, 3))
(If Amy sits on one of the 8 Y, Billy should take 1 of the 2 seats next to Amy,and the other three take 3 of the rest 8 seats.)
(No. of ways in the above situation = P(8, 1) x P(2, 1) x P(8, 3)

No. of ways if Amy and Billy sit next to each other
= P(2,1) x P(1, 1) x P(8, 3) + P(8,1) x P(2, 1) x P(8, 3)
= 2 x 1 x 8!/5! + 8 x 2 x 8!/5!
= 18 x 8!/5!

The required probability
= 18/90
= 1/5


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Method 2 :
No. of ways if 5 students sit randomly
= P(10,5)
= 10!/5!

(Consider the ways that Amy and Billy sitting next to each other.)
(Treat Amy and Billy as 1 student to take the seat. That means 9 students totake 4 seats)
(No. of ways in the above situation = P(9, 4)
(No. of ways of internal permutation = P(2, 2)

No. of ways if Amy and Billy sit next to each other
= P(9, 4) x P(2, 2)

The required probability
= P(9, 4) x P(2, 2) / P(10, 5)
= (9!/5!) x 2! / (10!/5!)
= (9!/5!) x 2 / (10 x 9!/5!)
= 2/10
= 1/5

Wish it helps!


圖片參考：http://imgcld.yimg.com/8/n/HA00132885/o/701203070016613873446510.jpg

五個亂坐的方法數目 = 10 P 5
Amy 同 Billy 坐埋一齊的方法數目 = 9 P 4 x 2 P 2
所以概率 = ( 9 P 4 x 2 P 2 ) / ( 10 P 5 ) = 1 / 5