Probability problem

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Darya

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What is the probability that a child, his parent and his parent's parent have the same zodiac sign?
I tried to solve it with probability trees. The probability that child's parent has the same sign is 1/12. His second parent has to have a different zodiac sign =. Following this logic, it's: [MATH]2*11/12*1/12(2*1/12*11/12)[/MATH]
Is it correct?
 
You're overthinking it.

A child has to be born in 1 out of 12 months.
The probability that the parent is born in this same month is 1/12.
The same goes for a grandparent.
 
What makes you say that exactly one parent can have the same sign as their child (in this problem)
 
Thinking along your lines (which I do not agree with) you have that the son can be born in any month with 100% certainty.

P(father, but not mother, being born on the same month as the son) = p(mother, but not father, being born on the same month as the son) = (1/12)(11/12). So p(exactly one parent having the same birthday as the son) = 2(1/12)(11/12) = 11/72.

Same goes for the grand parents. p( exactly one grandparent having the same birthday as the son) = 11/72.

p( exactly one parent AND exactly one grandparent having the same birthday as the son) = 1*(11/72)*(11/72) which is what you got. Good job doing that but I still wonder about your assumption.


EDIT: I forgot to include two sets of grandparents. So exactly one can be born on the same month as the son and that probability is (1/12)(11/12)^3. But this can happen in two ways--one from either set of grandparents. Hence the p(exactly one grandparent being born on the same month as the son) =2(1/12)(11/12)^3. Final answer is 22(1/12)(11/12)^3.*11/72
 
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Darya said:
Excuse me, I should've mentioned it in the problem
Click to expand...

So you're saying that one parent has the same sign and the other doesn't,
and then one grandparent has the same sign and the other 3 .. don't?
 
firemath said:
Good question. Maybe ask Subhotosh.:devilish:

What do you mean, "what?"
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What means what are you saying in the post? The OP had 1/12 all over the place. Given the way the OP interrupted the problem do you not thing that 11/12 needs to be there as well?
 
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Darya said:
Excuse me, I should've mentioned it in the problem
Click to expand...
Are you saying that it did say that? Can you post the entire problem exactly as given to you?

I must be tired because I forgot that one has two sets of grandparents and did not include that in my work. I will edit that.
 
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It seems to me that so far it is assumed that a child has the same probability of being born in February as in March? But then what about in a leap year?
 
pka said:
It seems to me that so far it is assumed that a child has the same probability of being born in February as in March? But then what about in a leap year?
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You made a good point, but do you really think that the author meant for the student to take into account how many days are in each month and then do it all over again for leap years?
 
Jomo said:
if you stayed in grade school longer you would have learned even more. I keep telling you that education is important but you never learn.
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It really was just a random question my friend asked me
It has to be in the way that it's one link. So if a child, his parent and this parent's parent has the same sign, all others cant have it. Disregard leap year.
 
Darya said:
It really was just a random question my friend asked me
It has to be in the way that it's one link. So if a child, his parent and this parent's parent has the same sign, all others cant have it. Disregard leap year.
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Disregarding leap year does not remove pka's point. The probability of being born in January is not the same as the probability of being born in April as they do not have the same number of days. p(being born in January) = 31/365 \(\displaystyle \neq \) 1/12 and p(being born in April) = 30/365 \(\displaystyle \neq\)1/12

One also has to assume, incorrectly, that the season does not matter. Many people choose to have their children in the winter so that when they are ready to go outside for walks...etc, it will be summer.
 
Actually, the question wasn't about months at all, so who cares about leap years or February? It was about zodiac signs, which I suppose we can presume to each cover exactly 1/12 of the year (at least for the sake of the problem). They are not tied to the beginning and end of months.

Of course, the assumption that births are uniformly distributed over the year is necessary if we don't want to collect statistics.

The bigger issue is what the event should be, which depends on why the question is asked. For example, if it were asked in order to be amazed at how unlikely it is that, in your family, a person in each generation has the same sign, I would think the relevant event should include even more unlikely events, such as everyone in the family having the same sign. Then the event ought to be "at least one person in each generation has the same sign": at least one child, at least one parent, and at least one grandparent".

But that's harder to work out, so I'd stick with the question as asked, which apparently is "a child, exactly one parent, and exactly one grandparent".
 
Jomo said:
Thinking along your lines (which I do not agree with) you have that the son can be born in any month with 100% certainty.

P(father, but not mother, being born on the same month as the son) = p(mother, but not father, being born on the same month as the son) = (1/12)(11/12). So p(exactly one parent having the same birthday as the son) = 2(1/12)(11/12) = 11/72.

Same goes for the grand parents. p( exactly one grandparent having the same birthday as the son) = 11/72.

p( exactly one parent AND exactly one grandparent having the same birthday as the son) = 1*(11/72)*(11/72) which is what you got. Good job doing that but I still wonder about your assumption.


EDIT: I forgot to include two sets of grandparents. So exactly one can be born on the same month as the son and that probability is (1/12)(11/12)^3. But this can happen in two ways--one from either set of grandparents. Hence the p(exactly one grandparent being born on the same month as the son) =2(1/12)(11/12)^3. Final answer is 22(1/12)(11/12)^3.*11/72
Click to expand...
Thank you! Exactly, you found my mistake. I missed that pair of grandparents too. Hope you're having a nice day!
 
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