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Tue 9 Jan, 2007 01:08 pm
Hi all, I wonder if i could get help with this problem on probability here.
Questions:
1.
A die is loaded in such a way that the probability of each face turning up is proportional to the number of dots on that face. e.g A six is three times as probable as a two. What is the probability of getting an even number in one throw?
2. You offer 3:1 odds that your friend Smith will be elected Mayor of your city. What probability are you assigning to the event that Smith wins?
I'll be much obliged with any responses. thanks in advance.
Time for homework, eh?
Why don't you tell us your thoughts, and we'll discuss.
I'll bet $30 to your $10 that no one here answers your homework for you..
What probability is it that you will get the answers?
parados wrote:I'll bet $30 to your $10 that no one here answers your homework for you...
Drewdad asks slyly, "Can I get a little of that action?"
I would have to say the odds of a post in this thread not mocking someone else are about the same as rolling a 1 on a die loaded in such a way that the probability of each face turning up is proportional to the number of dots on that face.
For those of you keeping score at home that would be odds of 1 in the summation of 1 to 6
shewolfnm wrote:should I sell tickets ?
Is there a chance that you will?
parados wrote:I'll bet $30 to your $10 that no one here answers your homework for you..
What probability is it that you will get the answers?
Probably that looks like a homework to you but it aint. Im actually trying to help a friend who is having problems as im not able to help. Besides im not in high school to be posting my homework on forums for solutions. You'll do yourself good by thinking before you post reply to threads. If i cant get help here no probs...its not the end of the world. definitely not a homework. thanks
Hey sky,
I would say the odds are good there might be hints in a couple of the posts already, probably better than 50/50 and maybe even better than the odds of getting an even number on the die in the first problem but certainly better than the odds of not getting an even number.
1. Think of it as there is one way to get a one, two ways to get a two, ..., six ways to get a six. What's the total for all six numbers? What's the total for the even numbers?
2. Odds are related to probability like this: if the odds are m:n the probability is m/(m+n).
Sounds like a betting question. Playing any dice lately? Maybe craps?
1.Hint:
P1..1/21 = 1/(1+2+3+4+5+6)
P2..2/21
P3..3/21
P4..4/21
P5..5/21
P6..6/21
(2.markr has already given a good hint.)
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