# Maths & Statistics question; probability.

• Dec 5th 2008, 06:01 AM
JohnT
Maths & 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?

• Dec 5th 2008, 10:37 AM
nzmathman
$P(boy) = \frac{33}{41}$

The probability of first two children being boys is $\frac{33}{41} \times \frac{33}{41}\,$ assuming each child arrives independently.
• Dec 5th 2008, 10:52 AM
JohnT
Quote:

Originally Posted by nzmathman
$P(boy) = \frac{33}{41}$

The probability of first two children being boys is $\frac{33}{41} \times \frac{33}{41}\,$ assuming each child arrives independently.

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 AM
nzmathman
I said $\frac{33}{41} \times \frac{33}{41}$ is the answer,

ie $\left(\frac{33}{41}\right)^2$
• Dec 5th 2008, 12:08 PM
mr fantastic
Quote:

Originally Posted by JohnT
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?

If you draw a tree diagram it should be clear that the answer is $\left(\frac{33}{41}\right) \cdot \left(\frac{32}{40}\right)$.

Quote:

Originally Posted by nzmathman
I said $\frac{33}{41} \times \frac{33}{41}$ is the answer,

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

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 PM
nzmathman
Oops, sorry about that mistake, should have realised there was no replacement (Doh)
• Dec 10th 2008, 07:35 AM
Johnt447
Looks like I got a clone lol