求兩個二重積分的比。Twfx is right. 5/9

令boy到達時間x，girl到達時間y， x 和y都可從0變到60。

所有可能性是二重積分x 和y都從0變到60。
S1 = integral|0,60 integral|0,60 dy dx = 60*60= 3600

相差20分鍾之內可能性是三個二重積分之和，
1.如boy到達時間從0到20，girl最早到達時間為0，最晚比boy晚到20。
If 0 <= x <= 20, then 0 <= y <= x+20;
2.如boy到達時間從20到40，girl最早比boy早到20，最晚比boy晚到20。
If 20 <= x <= 40, then x-20 <= y <= x+20;
3.如boy到達時間從40到60，girl最早比boy早到20，最晚到達時間為60。
If 40 <= x <= 60, then x-20 <= y <= 60;
S2 = integral|0,20 integral|0,x+20 dy dx +
integral|20,40 integral|x-20,x+20 dy dx +
integral|40,60 integral|x-20,60 dy dx
= integral|0,20 (x+20) dx + integral|20,40 (40) dx + integral|40,60 (80-x) dx
= (½*x^2+20*x)|0,20 + (40*x)|20,40 + (80*x - ½*x^2)|40,60
= (½*20^2+20*20) + (40*(40-20)) + (80*(60-40) - ½*(60^2-40^2))
= 600 + 800 + (1600 - ½*2000)
= 600 + 800 + 600
= 2000

The probability that they will actually meet is
S2 / S1 = 2000 / 3600 = 5/9.

所有跟帖： 

• Brother: U& Twfx too smart ,I do it by tipical math. -jinjing- ♀ (253 bytes) () 04/16/2010 postreply 08:12:56

• good! different ways to the same destination. -皆兄弟也- ♂ (0 bytes) () 04/16/2010 postreply 08:28:42