Question

Probability: There are 3 blue disks, 5 green disks, and nothing else in a container. Disks will be chosen one at a time, and at random and without replacement, from the container, until all 3 blue disks have been chosen. What is the probability that the third blue disk will be the seventh disk chosen?

Answer 1

since there are only 2 colors to play with a binomial sounds appropriate

Answer 2

but then maybe not ...

0 0 0 0 1 1 1
0 0 0 1 0 1 1
0 0 1 0 0 1 1
0 1 0 0 0 1 1
1 0 0 0 0 1 1

0 0 0 1 1 0 1
0 0 1 0 1 0 1
0 1 0 0 1 0 1
1 0 0 0 1 0 1

0 0 1 1 0 0 1
0 1 0 1 0 0 1
1 0 0 1 0 0 1

0 1 1 0 0 0 1
1 0 1 0 0 0 1

1 1 0 0 0 0 1

that might be all of the ways :)

Answer 3

3! 5! / 8! if you can determine the number of ways to pull them, the basic probability would resemble that

Answer 4

For the 1st 6 choices, only 2 blue are allowed so possible sample would be: ggggbb
There are \(\dbinom{6}{2}\) possible combinations of 4 green and 2 blue with each sample having equal probability: \(\dfrac{5}{8}\dfrac{4}{7}\dfrac{3}{6}\dfrac{2}{5}\dfrac{3}{4}\dfrac{2}{3}\).
The 7th choice being b having the probability, \(\dfrac{1}{2}\).
So total, this will be \(\dbinom{6}{2}\dfrac{5}{8}\dfrac{4}{7}\dfrac{3}{6}\dfrac{2}{5}\dfrac{3}{4}\dfrac{2}{3}\) = \(\dbinom{6}{2}\dfrac{5!3!}{8!}\), confirming @amistre64 's result.

Answer 5

thanks ! @ybarrap , however I am having trouble understanding where the \[\left(\begin{matrix}6 \\ 2\end{matrix}\right)\] comes from . I get the probabilities. Since there are 3 blue discs I get that we must account for any 2 being in the first 6 as well as any combination of 4 green discs in the first 6 . , but I would have done \[\left(\begin{matrix}3 \\ 2\end{matrix}\right)*\left(\begin{matrix}5 \\ 4\end{matrix}\right)\]

Answer 6

These two are equivalent: choosing 2 from 6 without replacement is the same as choosing 2 of 3 without replacement times the number of ways of choosing 4 from 5 w/o replacement. Depends on how you like to think about these things. You could have also said choose 4 out of 6 w/o replacement and this would have also given the same number of combinations. They all come out to 15.

Answer 7

Another way to look at \( \dbinom{6}{2}\) is to imagine how 2 heads can be arranged amongst 6 positions or 4 tails can be arranged among 6 positions, if heads=1 and tails=0.

However, they way you looked at the problem, you could separate the 6 positions into two groups, one with 2 members and the other with 4. There are 3 heads to choose for those 2 members and there are 5 tails to choose for those 4 remaining. So you're looking at the potential pool from each group and how you can arrange them into 2 and 4 respectively:

In my case, I didn't look at the potential pool, i just looked at what was possible with 2 heads and 6 places. In our case you looked outside the observations that there were two positions to arrange and thought in terms of 2 individual groups selected from a larger population and how they can be arranged.

Sometimes one way has an advantage over the other, but in this case there isn't a winner I think.