14 Apr 08
Originally posted by Green Paladinnice... because of the consecutive placement of the bullets, there are two chambers that if were chosen at random from the beginning, would allow for 3 consecutive empty chambers. so the odds of getting a bullet in the first 3 pulls is 2/6 or 1/3
[/i]1. Don't spin. There is only one chamber that will have a bullet in it next and three that will be empty. So these 1 in 4 odds are less than the initial 1 in 3 chance.
2. No difference. The odds are 1 in 3 whether you spin or not.
I know the answer to the boy/girl problem.
but of course this is the probability of getting a bullet with a random spin (2 bullet / 6 possible chambers). things change a little if you have to pull the trigger again. lol.
14 Apr 08
Originally posted by geepamoogle1) If the oldest child is a boy, then the chance that the second child is a girl is 1/2.
Suppose a family has two children. The children are not twins, and each child is either a boy or a girl, and equally likely to be either.
What are the chances the family has a girl if...
[b]1) the oldest is a boy?
2) they have at least one boy?[/b]
2) If the family has at least one boy, then the chance that the other child is a girl is 2/3.
14 Apr 08
In the Russian Roulette problem, you would want to spin if the bullets were paced in nonconsecutive chambers.
If they were placed completely randomly, I don't know what the answer would be yet.
And you are correct PBE6.
(The paradox there is in the fact that some people think the two questions are the same. The first one includes extra information.)
14 Apr 08
Originally posted by geepamoogleI think these questions are great. They're simple enough to grasp quickly, but the answers make you think just a little bit longer about the problem. That's the hook that gets people learning.
In the Russian Roulette problem, you would want to spin if the bullets were paced in nonconsecutive chambers.
If they were placed completely randomly, I don't know what the answer would be yet.
And you are correct PBE6.
(The paradox there is in the fact that some people think the two questions are the same. The first one includes extra information.)
14 Apr 08
Originally posted by deriver69wrong. Read my above explanation.
It does make a difference. The four possible outcomes are like the four possible outcomes with 2 coins being tossed. (HH HT TH or TT)
Out of the four possible outcomes three of them have at least one boy. two of these three have a girl.
14 Apr 08
1 edit
Originally posted by eldragonflyYes that is exactly what I am saying. Put another way
Doesn't make a difference in this problem. Stated another way what you're actually implying is that there is only a 1/3 chance that the other child is a boy, given that one child is a boy.
P(other child is a girl|a child is a boy)=P(one is a girl the other is a boy)/P(at least one child is a boy)
P(other child is a girl|a child is a boy)=(2/4)/(3/4)=2/3
14 Apr 08
1 edit
Originally posted by eldragonflyHe's right, you know.
wrong. Read my above explanation.
Writing the eldest first, there are four possibilities with equal probability. BB, BG, GB, GG. They are equal in probability because we assume that the sex of the first child is independent of the second child.
The additional information rules out one possibility. So we have three possibilities..but still with equal probability. Two of the three match what we're looking for, so the probability is 2/3.
Stated another way what you're actually implying is that there is only a 1/3 chance that the other child is a boy, given that one child is a boy.
A 1/3 chance that there are two boys, given that we know there's at least one. Yes.
14 Apr 08
1 edit
Everytime I come back here there are another 50 or so comments. Have we done the one where if the four aces (2 red and 2 black) are face down on the table, what is the probability that two cards turned over will be of the same colour?
Actually I think it might be similar the boy/girl problem but it is getting late