STPM-学校功课討論区

dunwan2tellu

the letter in the word 'GANJIL' are rearranged. find the probability that the word formed randomly.
(a)begins with a consonant
  P(consonant)= 4/6 = 2/3

consonant = g,n,j,l
所以你开头有 4 个方法，剩下的就 5! 所以方法共有 4 x5!
Probability = (4 x 5!)/6! = 2/3

(b) ends with a vowel
  P(vowel) = 2/6 = 1/3

vowel = a,i
结尾有 2 个方法，剩下的就 5!
Probability = (2x5!)/6! = 1/3

(c) begins with a consonant or ends with a vowel.

你可以用 1 - P(not begin with consonant AND not end with vowel)
所以你开头能是 vowel , 结尾只能是 consonant .
方法有 2 x 4 x 4!
P(begin with consonant OR end with vowel)
= 1 - P(not begin with consonant AND not end with vowel)
= 1 - (2x4x4!)/6!
=11/15

原帖由 dunwan2tellu 于 31/7/2006 16:08 发表

consonant = g,n,j,l
所以你开头有 4 个方法，剩下的就 5! 所以方法共有 4 x5!
Probability = (4 x 5!)/6! = 2/3


vowel = a,i
结尾有 2 个方法，剩下的就 5!
Probability = (2x5!)/6! = 1/3



...

我明白啦！谢谢

还有麻烦啦！
a box contains 7 identical red balls 4 identical white balls and 9 identical black balls,3 balls are taken out from the box one after another without replacement. what is the probability that
(a)2 of them are red and one is black?
(b)all three balls have different colours?
(c)at least one ball is white?
(d) all three ahve the same colour?

还有麻烦啦！
a box contains 7 identical red balls 4 identical white balls and 9 identical black balls,3 balls are taken out from the box one after another without replacement. what is the probability that
(a)2 of them are red and one is black?
(b)all three balls have different colours?
(c)at least one ball is white?
(d) all three ahve the same colour?

还有麻烦啦！
a box contains 7 identical red balls 4 identical white balls and 9 identical black balls,3 balls are taken out from the box one after another without replacement. what is the probability that
(a)2 of them are red and one is black?
(b)all three balls have different colours?
(c)at least one ball is white?
(d) all three ahve the same colour?

dunwan2tellu

a box contains 7 identical red balls 4 identical white balls and 9 identical black balls,3 balls are taken out from the box one after another without replacement. what is the probability that
(a)2 of them are red and one is black?

2 red one black 方法有 7C2 x 9C1
total = 20C3 , P(2 red,,1 black) = (7C2 x 9C1)/(20C3) = ....

(b)all three balls have different colours?

表示 red , black ,white 各有一个所以方法有 7C1 x 4C1 x 9C1
P(1r,1b,1w) = (7C1 x 4C1 x 9C1)/(20C3) = ....

(c)at least one ball is white?

P(>= 1 W) = 1 - P(no white)
          = 1 - 16C3/20C3
          = ....

(d) all three ahve the same colour?

P(all same color) = P(all red) + P(all B) + P(all W)
                  = (7C3 + 4C3 + 9C3)/20C3
                  = ...

原帖由 dunwan2tellu 于 1/8/2006 07:14 发表

2 red one black 方法有 7C2 x 9C1
total = 20C3 , P(2 red,,1 black) = (7C2 x 9C1)/(20C3) = ....


表示 red , black ,white 各有一个所以方法有 7C1 x 4C1 x 9C1
P(1r,1b,1w) = (7C1 x 4C1 x 9C1)/(20 ...

原来是这样！那我想问为什么要用COMBINATION？

dunwan2tellu

原帖由 ~ciyun~ 于 1-8-2006 08:36 PM 发表

原来是这样！那我想问为什么要用COMBINATION？

因为你不需要知道“次序”。意思是说你先拿 black 再拿 white , 或是先 white 再 black ,都没有分别。同样都只是“拿两个”。在这个情况下，我们就用 C

如果需要“次序”，那么就要 P

nikuang04

这里有一题我学校的历年考试题目，我不会做！各位帮帮忙
Sin 3X = cos X
找X。

dunwan2tellu

这里有一题我学校的历年考试题目，我不会做！各位帮帮忙
Sin 3X = cos X

我们必须知道 x 的范围，不然就会有无限副答案．

一般来说我们可以用ｇｅｎｅｒａｌ　ｆｏｒｍｕｌａ , i.e

if sin x = k , then

x = Arcsin k = n * pi + (-1)^n * t  ,  t = principal value of sin x = k

if cos x = k , then
x = Arccos k = 2n * pi + - t , t = principal value of cos x = k

if tan x = k , then
x =Arctan k = t + - n * pi , t = principal value of tan x = k

So

if Sin 3x = Cos x = Sin(Pi/2 - x)

=> 3x = Arcsin(Sin(Pi/2-x)) = n * pi + (-1)^n * (Pi/2-x)

n = even ; 3x = n * pi + (pi/2 - x) <==> x = 1/4 * (n * pi + pi/2)
n = odd ; 3x = n*pi - (pi/2-x) <==> x = 1/2 * (n*pi - pi/2)

nikuang04

原帖由 dunwan2tellu 于 2-8-2006 21:48 发表


我们必须知道 x 的范围，不然就会有无限副答案．

一般来说我们可以用ｇｅｎｅｒａｌ　ｆｏｒｍｕｌａ , i.e

if sin x = k , then

x = Arcsin k = n * pi + (-1)^n * t  ,  t = principal value of  ...

我不知道你在做什么......
假设x是在和３６０
没有比较简单的方法吗?
什么principal ......根本不懂

dunwan2tellu

0 =< x =< 360 , 求 sin 3x = cos x 的解。

解答：sin 3x = cos x = sin(90 - x)
所以可以是 3x = 90 - x <==> x = 22.5

也可以是 sin 3x = cos x = sin(360 + 90 - x)
所以 3x = 450 - x <==> x = 112.5

或 sin3x = cos x = sin(720 + 90 - x)
3x = 810 - x <==> x = 202.5

或 sin 3x = cos x = sin (1080 + 90 -x)
3x = 1170 - x <==> x = 292.5

或 sin 3x = sin(90 + x)
3x = 90 + x <==> x = 45

或 sin 3x = sin(360 + 90 + x)
3x = 450 + x <==> x = 225

总共 6 副答案

nikuang04

原帖由 dunwan2tellu 于 2-8-2006 22:46 发表
0 =< x =< 360 , 求 sin 3x = cos x 的解。

解答：sin 3x = cos x = sin(90 - x)
所以可以是 3x = 90 - x <==> x = 22.5

也可以是 sin 3x = cos x = sin(360 + 90 - x)
所以 3x = 450 - x &l ...

厉害厉害
佩服佩服

可是为什么你会想到  sin 3x = cos x = sin(360 + 90 - x)
然后再一个一个家360呢？

最后第二个解答不是很明白
为什么  sin 3x =  sin(90 + x)

dunwan2tellu

原帖由 nikuang04 于 2-8-2006 10:51 PM 发表


厉害厉害
佩服佩服

可是为什么你会想到  sin 3x = cos x = sin(360 + 90 - x)
然后再一个一个家360呢？

因为我做过类是的题目。

最后第二个解答不是很明白
为什么  sin 3x =  sin(90 + x)

因为 sin(90+x) = cos x

只要是加减 90 ，270 的话，sin 会变去 cos ; cos 会变去 sin . 再来你只要看是 positive 还是 negative

几个基本的 identity (sin 的）
sin(90-x) = cos x
sin(90+x) = cos x
sin(180 - x) = sin x
sin(180 + x) = -sin x
sin(270 - x) = -cos x
sin(270+x) = -cos x
sin(360-x) = -sin x
sin(390+x) = sin x
sin(-x) = -sin x

nikuang04

原帖由 dunwan2tellu 于 2-8-2006 23:13 发表

因为我做过类是的题目。


因为 sin(90+x) = cos x

只要是加减 90 ，270 的话，sin 会变去 cos ; cos 会变去 sin . 再来你只要看是 positive 还是 negative

几个基本的 identity (sin 的）
sin(90-x ...

我一直以为只有 cos=sin (90-X)的
原来 cos X 也是 sin(90+x)
我的数学真的算不了什么

原帖由 dunwan2tellu 于 1/8/2006 20:44 发表


因为你不需要知道“次序”。意思是说你先拿 black 再拿 white , 或是先 white 再 black ,都没有分别。同样都只是“拿两个”。在这个情况下，我们就用 C

如果需要“次序”，那么就要 P

理解谢谢！

two dice are tossed together and the biggest of the two numbers abtained is recorded. if A represents the event'the biggest number is 4' and B represent the event 'the biggest number is odd' find
(a) P(A)
(b) P(A U B)
(c) P(A N B)

dunwan2tellu

two dice are tossed together and the biggest of the two numbers abtained is recorded. if A represents the event'the biggest number is 4' and B represent the event 'the biggest number is odd' find
(a) P(A)
(b) P(A U B)
(c) P(A N B)

先 list 出 outcome , i.e
A = (1,4),(2,4),(3,4),(4,4)，(4,1),(4,2),(4,3) ;
B = ( 1,1),(1,3),(3,1),(2,3),(3,2),(3,3),(1,5),(5,1),(2,5),(5,2),(3,5),(5,3),(4,5),(5,4),(5,5)
Total outcome = 6 x 6 = 36

然后你就从刚刚 list 出来的 outcome 来做。你先试试。

[ 本帖最后由 dunwan2tellu 于 5-8-2006 07:22 PM 编辑 ]

邵逸夫

a) log2 ( x+2 ) - log2 ( x-2 ) > 1
b) if a^2 and b^2 are the roots of x^2 -21x+4=0 and a and b are both positive, find
  i) ab and a+b
  ii) the equation with roots 1/a^2 and 1/b^2
c)find the range of values of the function y=(x^2 - x + 1)/ (x^2 + x + 1)
  for real values of x. hence state the maximum and minimum values of the function.

dunwan2tellu

原帖由 邵逸夫 于 7-8-2006 02:03 PM 发表
a) log2 ( x+2 ) - log2 ( x-2 ) > 1
b) if a^2 and b^2 are the roots of x^2 -21x+4=0 and a and b are both positive, find
  i) ab and a+b
  ii) the equation with roots 1/a^2 and 1/b^2
c)find the range of values of the function y=(x^2 - x + 1)/ (x^2 + x + 1)
  for real values of x. hence state the maximum and minimum values of the function.

a)找 x 吗？
首先必须满足 x+2>0 , 和 x-2> 0 ==> x>2 ...(1)
之后用log 的rule， log2 (x+2)/(x-2) > 1 ==> (x+2)/(x-2) > 2
==> (x+2)/(x-2) - 2 > 0 ==> (6-x)/(x-2) > 0
==> (x-6)(x-2) < 0
==> 2<x<6 ....(2)

(1),(2) => 2<x<6

b)sum of roots = a^2 + b^2 = 21
  product of roots = (ab)^2 = 4 ==> ab = 2 ( 因为 a,b >0)
所以 a + b = Sqrt[ a^2 + b^2 + 2ab ] = ....

(ii) sum of new roots = 1/a^2 + 1/b^2 = (a^2 + b^2)/(ab)^2 = k
    product of new roots = 1/a^2 * 1/b^2 = m

equation = x^2 - (k+m) + km = 0
k,m 你自己试试找。

c)y(x^2 + x + 1) = x^2 - x + 1
<==> (y-1)x^2 + (y+1) x + (y-1) = 0
to have real value , b^2 -4ac >= 0
所以 b^2 - 4ac = (y+1)^2 - 4(y-1)^2 >= 0 ==> (3y-1)(3-y) >= 0
==> 1/3 =< y =< 3
所以范围是 1/3 到 3 之间，minimum 和 maximum 也就不用我多说了。

[ 本帖最后由 dunwan2tellu 于 9-8-2006 02:18 PM 编辑 ]