Question

I've got a problem here: let's say we have an urn with 20 balls, 10 of which are white. What's the probability that at least one white ball was drawn after two successive draws(no replacement) and what formula is used to do the calculation?

Thank you for any replies.

Answer 1

This can be solved as follows:
1. Find the probability of drawing no white balls after two draws:
\[P(0\ white\ first\ draw)=P _{1}=\frac{10}{20}\]
\[P(0\ white\ second\ draw)=P _{2}=\frac{9}{19}\]
\[P(0\ white\ after\ 2\ draws)=P _{1} \times P _{2}=P _{3}=\frac{10}{20} \times \frac{9}{19}\]
2. The probability that at least one white ball was drawn after two successive draws(no replacement) is given by:
\[1-P _{3}=[1-(\frac{10}{20} \times \frac{9}{19})]=you\ can\ calculate\]

Answer 2

Thank you very much for your reply first. I get kinda the same result using combinations, so: the nr. of all combinations is 20 over 2 right and the way of picking two none white balls is 10 over 2 and then i subtract this from 1. Am I correct with this or have I made any mistakes?

Answer 3

You will get exactly the same result as given by the method that I posted if you use:
\[P(at\ least\ 1\ white)=1-\frac{\left(\begin{matrix}10 \\ 2\end{matrix}\right)}{\left(\begin{matrix}20 \\ 2\end{matrix}\right)}\]

Answer 4

Great, thank you.

Answer 5

You're welcome :)