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Mathematics 18 Online
OpenStudy (mathmath333):

If the first child of a couple is a boy.Find the probablity that the second child being a boy.

OpenStudy (mathmath333):

\(\large \color{black}{\begin{align} & \normalsize \text{ If the first child of a couple is a boy.Find the }\hspace{.33em}\\~\\ & \normalsize \text{probablity that the second child being a boy.}\hspace{.33em}\\~\\ \end{align}}\)

OpenStudy (anonymous):

lol

Parth (parthkohli):

Depending on one's interpretation of probability, the answer is either 1/2 or 1/3.

OpenStudy (alexandervonhumboldt2):

0? cause in a couple there is 1 girl and 1 boy if i remmeber correctly

OpenStudy (anonymous):

hahaha

OpenStudy (anonymous):

i think this type of question would not account for past results...

OpenStudy (anonymous):

unless your a baby predictor

Parth (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 ?

OpenStudy (alexandervonhumboldt2):

hahah gane, i'm bad at prob, but i would say 0

Parth (parthkohli):

@ganeshie8 What do you think the answer should be? 1/2?

OpenStudy (anonymous):

is that called mutually exclusive? i forgot

OpenStudy (mathmath333):

1/2

OpenStudy (alexandervonhumboldt2):

it can be 1/2 as well

OpenStudy (anonymous):

what if you got con-joined twins one a female and the other a male

OpenStudy (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

OpenStudy (mom.):

1/3 (;

Parth (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.

Parth (parthkohli):

But again, it should be 1/2.

OpenStudy (mathmath333):

It is from independent event problem

OpenStudy (alexandervonhumboldt2):

hmm bg=gb it is not said that the order is needed i would say 1/2 or 0

OpenStudy (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.

Parth (parthkohli):

Ohhh, so it's about an unborn baby.

Parth (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.

Parth (parthkohli):

If we're differentiating between the elder and the younger child, then 1/2 is surely the answer.

OpenStudy (mom.):

(;

OpenStudy (alexandervonhumboldt2):

\(\Huge\color{black}{☺☻☺☻☺☻☺☻☺☻}\)

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