mathmath333
  • mathmath333
If the first child of a couple is a boy.Find the probablity that the second child being a boy.
Mathematics
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SOLVED
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chestercat
  • chestercat
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} & \normalsize \text{ If the first child of a couple is a boy.Find the }\hspace{.33em}\\~\\ & \normalsize \text{probablity that the second child being a boy.}\hspace{.33em}\\~\\ \end{align}}\)
anonymous
  • anonymous
lol
ParthKohli
  • ParthKohli
Depending on one's interpretation of probability, the answer is either 1/2 or 1/3.

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AlexandervonHumboldt2
  • AlexandervonHumboldt2
0? cause in a couple there is 1 girl and 1 boy if i remmeber correctly
anonymous
  • anonymous
hahaha
anonymous
  • anonymous
i think this type of question would not account for past results...
anonymous
  • anonymous
unless your a baby predictor
ParthKohli
  • ParthKohli
Read this: http://mathforum.org/dr.math/faq/faq.boygirl.choose.html
ganeshie8
  • ganeshie8
similar problem : If you rolled a coin two times and if you saw HEAD first time, whats the probabiltiy that you see a HEAD on second roll too ?
AlexandervonHumboldt2
  • AlexandervonHumboldt2
hahah gane, i'm bad at prob, but i would say 0
ParthKohli
  • ParthKohli
@ganeshie8 What do you think the answer should be? 1/2?
anonymous
  • anonymous
is that called mutually exclusive? i forgot
mathmath333
  • mathmath333
1/2
AlexandervonHumboldt2
  • AlexandervonHumboldt2
it can be 1/2 as well
anonymous
  • anonymous
what if you got con-joined twins one a female and the other a male
anonymous
  • anonymous
if thats ever possible
ganeshie8
  • ganeshie8
Yes, it has to be 1/2 if it is an unbiased coin; the coin has no memory, it wont remember the previous roll and decide itself to roll on a different face the next time
mom.
  • mom.
1/3 (;
ParthKohli
  • ParthKohli
That's what I think too. However, this problem is a lot like the Monty Hall Problem, and given that one child is a boy, we eliminate the GG case, which leaves us with BG, GB, BB. That sounds like 1/3.
ParthKohli
  • ParthKohli
But again, it should be 1/2.
mathmath333
  • mathmath333
It is from independent event problem
AlexandervonHumboldt2
  • AlexandervonHumboldt2
hmm bg=gb it is not said that the order is needed i would say 1/2 or 0
mom.
  • mom.
female gametes-->XX male gametes --->XY when u fuse u get either ----> XX or XY so it will be 1/2
ganeshie8
  • ganeshie8
The probability would be different only if the experiment had already taken place, I think. Then, we can use Bayee inference and use conditional probability..
ganeshie8
  • ganeshie8
If the second coin is not rolled yet, then the probability of seeing HEAD on it is 1/2; this much is non negotiable.
ganeshie8
  • ganeshie8
On the other hand, if you have rolled the coins and had the results already, then the sample space is : {HH, HT, TH, TT} the probability would be 1/3; but clearly, in the main question the baby was not born yet, so the probabiblity for our main question should be 1/2.
ParthKohli
  • ParthKohli
Ohhh, so it's about an unborn baby.
ParthKohli
  • ParthKohli
Then definitely 1/2.
ganeshie8
  • ganeshie8
** On the other hand, if you have rolled the coins and had the results already, then the sample space is : {HH, HT, TH, TT} If somebody tells you that one coin is H, the probability for other coin to be H would be 1/3; but clearly, in the main question the baby was not born yet, so the probabiblity for our main question should be 1/2.
ParthKohli
  • ParthKohli
If we're differentiating between the elder and the younger child, then 1/2 is surely the answer.
mom.
  • mom.
(;
AlexandervonHumboldt2
  • AlexandervonHumboldt2
\(\Huge\color{black}{☺☻☺☻☺☻☺☻☺☻}\)

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