In a class of 33 boys and 8 girls what is the probability of the first and second child to arrive in the morning to be a boy?

Thanks in advance to anybody who answers(Happy)

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- Dec 5th 2008, 06:01 AMJohnTMaths & Statistics question; probability.
In a class of 33 boys and 8 girls what is the probability of the first and second child to arrive in the morning to be a boy?

Thanks in advance to anybody who answers(Happy) - Dec 5th 2008, 10:37 AMnzmathman
$\displaystyle P(boy) = \frac{33}{41}$

The probability of first two children being boys is $\displaystyle \frac{33}{41} \times \frac{33}{41}\,$ assuming each child arrives independently. - Dec 5th 2008, 10:52 AMJohnT
Hey, thanks for taking the time to reply.

I am not big on Maths, and am doing this as a small part of my Sociology course (Statistics), but the answer you gave is exactly the same answer that I gave, 33/41. I thought it might be wrong when I wrote it but wanted to write something rather than nothing.

It is marked wrong. So are we sure that is the right answer? - Dec 5th 2008, 11:10 AMnzmathman
I said $\displaystyle \frac{33}{41} \times \frac{33}{41}$ is the answer,

ie $\displaystyle \left(\frac{33}{41}\right)^2$

- Dec 5th 2008, 12:08 PMmr fantastic
If you draw a tree diagram it should be clear that the answer is $\displaystyle \left(\frac{33}{41}\right) \cdot \left(\frac{32}{40}\right)$.

This answer will only be correct if the first child goes back home because s/he forgot his/her lunch .....

(In other words, sampling with replacement) - Dec 5th 2008, 12:48 PMnzmathman
Oops, sorry about that mistake, should have realised there was no replacement (Doh)

- Dec 10th 2008, 07:35 AMJohnt447
Looks like I got a clone lol