Probability

2007-06-14 9:50 pm
There are 5 lines segments with lengths 1cm,3cm,5cm,7cm and 9cm respectively. If 3 of them are chosen at random, what is the probability that they can form a triangle?

要詳盡列式..唔該

回答 (3)

2007-06-14 10:10 pm
✔ 最佳答案
Total combination=5 x 4 x 3=60

To form an triangle => the sum 0f two short lines >= the long one

If we draw 1 cm out, then the combintaion to form an triangle is 0 (since 1+3=4, 1+5=6, 1+7 = 8, none of these is longer than the longest one)

If 3 cm is draw, the combinations are [(3,5,7) ,(3,5,9) and (3,7,9)]

We can see all of the rest of number are listed, we need to calculate the no. of combinations.

For 3,5,7 , there is 3 x 2 x 1 combinations. (3,5,9), (3,9,5),(5,9,3),(5,3,9),(9,3,5),(9,5,3)
For 3,5,9 , there is 3 x 2 x 1 combinations.
For 3,7,9 , there is 3 x 2 x 1 combinations.

Total = (3 x 2 x 1) + (3 x 2 x 1) + (3 x 2 x 1) =18

probability = 18/60=0.3
2007-06-14 10:09 pm
可能性係 3/10

首先, 一共有 5C3 = 10 個可能性
基本上三條線可以 form 到一個三角形, 就係兩條邊加埋一定要比第三條邊(最長嗰條)長
以 5 作為最長邊, 已經 form 唔到三角 (因 1+3<5)
以 7 作為最長邊, 只能由 3, 5 & 7 組成三角
以 9 作為最長邊, 可以分別由 3, 7 & 9 或 5, 7 & 9 組成三角

即係只有三個組合, ie 3,5,7 / 3,7,9 / 5,7,9 可以組成三角形...
(不過唔知有冇真係計出黎既方法呢~)
2007-06-14 10:03 pm
依照你既問題,應該係100%
因為任何長度既三條line segment都可以from 到 triangle


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