I'm trying to teach myself probability, but it's really difficult for me to understand.
I'm trying this question from a textbook I've downloaded online, and I just can't seem to wrap my head around it.
Question 1. a) There is a family of 3 children. You saw one of the children, and it's a girl. What's the probability that all of them are girls?
1. b) You saw another child. You're not sure if it's the same girl you saw. What's the probability that all of the children are girls? Is it the same as question 1. a)?
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Anonymous2011-05-29 20:08
OP here.
I'm thinking 1/3 for the 1. a)
I'm not sure what 1. b) is, though.
Any ways I can approach this problem?
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Anonymous2011-05-30 4:44
1.a) The answer is the probability that the two other children are girls, which is not 1/3.
1.b) It's not the same question hwre because you have two cases you need to distinguish: the case where it is the same girl and the case where it is not.
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Anonymous2011-05-30 6:54
This is just conditional probability.
For (a) your sample space can be all possible single drawings from \{B,G\}^3.
For (b) your sample space can be all possible double drawings (with replacement) from \{B,G\}^3.
Your answer to (a) is wrong assuming a uniform sample space (i.e., each child was equally likely to be seen).
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Anonymous2011-05-30 15:42
a- 0.5X0.5X1.
Probability for a child being a female is 0.5. For the second one too. The third one is already a known so it's 1.
b I lold
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Anonymous2011-06-05 16:50
This all depends on how insane you want to get on accuracy. There's a slight difference in male:female ratio, (although it's not clear on if that's by virtue of other factors, or differing birth rate. Someone might know, I don't.). For the scope of the question, I'll assume it's 50:50.
As for b., well, yes it's the same question, unless you want to take the one female child into account based on how many female children there are in the world and alter the probability of the other 2 being female by an infinitesimally small amount.
</pedant>
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Anonymous2011-06-05 17:12
This all depends on how insane you want to get on accuracy. There's a slight difference in male:female ratio, (although it's not clear on if that's by virtue of other factors, or differing birth rate. Someone might know, I don't.). For the scope of the question, I'll assume it's 50:50.
As for b., well, yes it's the same question, unless you want to take the one female child into account based on how many female children there are in the world and alter the probability of the other 2 being female by an infinitesimally small amount.
</pedant>
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Anonymous2011-06-05 17:13
>>7 >>6
My apologies for the double posting - I don't frequent the BBS, and I'm posting from my phone. Sorry.
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Anonymous2011-06-09 18:41
Ok, are you leaving something out of problem b? Because the way I read it is like this: You see one girl in the family of three. Then you see some other child. No information is given about whether or not that child is in the family or about gender. Unless we're calling the observer effect, I'm not sure how it's even relevant that you just saw some child.
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Anonymous2011-06-25 11:46
Not sure if it's the girl you saw first =2/3 chance there's another girl. then there's the one you dont know about = 0,5
2/3 * 0,5 = 0,33 ?
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Anonymous2011-06-29 0:02
A: Yeah, 3 Persons, 1 girl (100%) = 1, 2 Persons not sure (50%) = 0,5X0,5 = 0,5² so multiplie all possibilities like u always do; 1X0,5²= 0,25
B: you have to set the priority of the information you get right.
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oddSpace2011-06-30 11:42
For the first question, 1/7
A family having three children has a 1/8 probability of getting three girls
and a 1/8 probability of not having *any* girls
Therefore if one of them is known to be female and you have no information on the others then we can eliminate the situation in which they are all male, leaving seven possibilities. Only one of these is that all three are girls. There is a 1/7 chance.
The second is more complex. There's a 2/3 chance that the new person is a different girl, and therefore the three situations in which two children are male can be eliminated, leaving four remaining. This means there is a 2/3 chance that there is a 1/4 chance.
there's a 1/3 chance that you have no new information. This will have a 1/7 chance.
This means your total probability is :
(2/3)*(1/4)
+
(1/3)*(1/7)
= 1/6 + 1/21
=27/126
I probably got something unimportant like the actual numbers wrong, but that's pretty much it.
en.wikipedia.org/wiki/boy_or_girl_paradox
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Anonymous2011-06-30 18:00
>>9
Presumably the child is from the same family. Also presumably you can tell the difference between boys and girls.
>>12
It's 1/7 only if you are biased and see girls whenever possible, in which case a second girl sighting makes no difference. And "a 2/3 chance that the new person is a different girl" is wrong, as that only holds when all the children are girls. In general a repeat is more common than that.
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Anonymous2011-07-01 4:41
the lower the probability, the higher the impossible is likely to happen.
Calculating probability is useful for checking solutions to problem and number puzzles.
In math, probability is expressed as a 0 (impossible) or 1 (certain) with probable outcomes expressed as a number between 0 and 1, represented as a fraction.
For example, if you toss a coin once, it can only land heads or tails - the chance of it landing heads is therefore one in two, expressed as 1/2. You would express the chance of rolling a 5 on a six-sided die as 1/6 . But if you are looking to roll either a 5 or a 3 in one roll, your chances improve to 1/6 + 1/6, which makes 1/3. However, the probability of the president being in Denver and San Francisco at the same time would be 0 as it is impossible.