# Thread: Maths & Statistics question; probability.

1. ## 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?

Thanks in advance to anybody who answers

2. $\displaystyle P(boy) = \frac{33}{41}$

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

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

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

4. I said $\displaystyle \frac{33}{41} \times \frac{33}{41}$ is the answer,

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

5. 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?

Thanks in advance to anybody who answers
If you draw a tree diagram it should be clear that the answer is $\displaystyle \left(\frac{33}{41}\right) \cdot \left(\frac{32}{40}\right)$.

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

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

6. Oops, sorry about that mistake, should have realised there was no replacement

7. Looks like I got a clone lol