 |
|
| Science Forum Index » Mathematics Forum » high school....probability... |
|
Page 1 of 1 |
|
| Author |
Message |
| mina_world |
 Posted: Thu Nov 25, 2004 1:59 am |
|
|
|
Guest
|
hello......doctor~
Two dice are rolled at once.
Determine the probability that
the number of one dice is the multiple of number of other dice.
------------------------------------------------
1)
all event is
(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)
(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)
(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)
(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)
(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)
(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)
so, multiple events is
(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)
(2,2)(2,4)(2,6)
(3,3)(3,6)
(4,4)
(5,5)
(6,6)
and
(2,1)(3,1)(4,1)(5,1)(6,1)
(4,2)(6,2)
(6,3)
thus, 22 / 36 = 11 / 18
2)
all event is
(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)
(2,2)(2,3)(2,4)(2,5)(2,6)
(3,3)(3,4)(3,5)(3,6)
(4,4)(4,5)(4,6)
(5,5)(5,6)
(6,6)
so, multiple events is
(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)
(2,2)(2,4)(2,6)
(3,3)(3,6)
(4,4)
(5,5)
(6,6)
thus, 14 / 21 = 2 / 3
------------------------------------------
um.....both (1) and (2) possible solutions ??
but the answer is (1) by answer paper.
and the mention about size and color of two dice doest not exist.
method (2) is really wrong ?
how do you think about it ?
thank you very much for your advice. |
|
|
| Back to top |
|
|
|
| Ron Sperber |
| Posted: Thu Nov 25, 2004 8:03 am |
|
|
|
Guest
|
mina_world wrote:
[quote:bcc2cda9b8]hello......doctor~
Two dice are rolled at once.
Determine the probability that
the number of one dice is the multiple of number of other dice.
------------------------------------------------
1)
all event is
(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)
(2,1)(2,2)(2,3)(2,4)(2,5)(2,6)
(3,1)(3,2)(3,3)(3,4)(3,5)(3,6)
(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)
(5,1)(5,2)(5,3)(5,4)(5,5)(5,6)
(6,1)(6,2)(6,3)(6,4)(6,5)(6,6)
so, multiple events is
(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)
(2,2)(2,4)(2,6)
(3,3)(3,6)
(4,4)
(5,5)
(6,6)
and
(2,1)(3,1)(4,1)(5,1)(6,1)
(4,2)(6,2)
(6,3)
thus, 22 / 36 = 11 / 18
2)
all event is
(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)
(2,2)(2,3)(2,4)(2,5)(2,6)
(3,3)(3,4)(3,5)(3,6)
(4,4)(4,5)(4,6)
(5,5)(5,6)
(6,6)
so, multiple events is
(1,1)(1,2)(1,3)(1,4)(1,5)(1,6)
(2,2)(2,4)(2,6)
(3,3)(3,6)
(4,4)
(5,5)
(6,6)
thus, 14 / 21 = 2 / 3
------------------------------------------
um.....both (1) and (2) possible solutions ??
but the answer is (1) by answer paper.
and the mention about size and color of two dice doest not exist.
method (2) is really wrong ?
how do you think about it ?
thank you very much for your advice.
Yes, method 2 is really wrong. The problem is that in method 2 you are[/quote:bcc2cda9b8]
discounting order. The problem with that is that discounting order
if a and b are different numbers then the pair (a,b) is twice as likely
as a pair of the form (a,a). But in method 2 you are treating the
_unordered pair_ (2,4) as equally likely as say (4,4) which is incorrect. |
|
|
| Back to top |
|
|
|
|
|
All times are GMT - 5 Hours
The time now is Wed Dec 02, 2009 8:59 pm
|
|