there are two urns with White balls and Black balls. first urn has 21 whites and 5 blacks, second one has 8 whites and 9 blacks. we take 7 balls from first urn and put them into the second one. afterwards, out of the second urn we take one ball. what is the probability that it's white?
I've been struggling on this probability for quite some time now and eventually got overwhelmed by numbers and every single way the balls can be taken from one urn to another, any help will be appreciated
On
Method: Calculate for every white ball the probability to become the chosen ball and add these probabilities (of mutual exclusive events).
A white ball that was at the start of this in the first urn has probability $\frac{7}{26}\frac{1}{24}$ to be chosen and there are $21$ such balls.
A white ball that was at the start of this in the second urn has probability $\frac{1}{24}$ to be chosen and there are $8$ such balls.
So the probability that a white ball will be chosen is: $$21\times\frac{7}{26}\frac{1}{24}+8\times\frac{1}{24}$$
edit:
The answer of P. Lawrence made me aware that initially I made a mistake.
That is now repaired: $\frac1{26}$ is replaced by $\frac7{26} $.
Sorry if I caused any confusion.
On
The probability that any pariular ball in the first urn is moved to the second urn is $\frac{{1 \choose 1} \times {25 \choose 6}}{{26 \choose 7}}=\frac {7}{26}$.The probability that that particular ball is then chosen from the second urn is $\frac{1}{24}$. So the probability that a particular ball that was originally in the second urn is finally chosen is $\frac{7}{24 \times 26}.$ There are 21 white balls in the first urn so the probability of one of them being chosen is $\frac{21 \times 7}{24 \times 26}.$ The probability of a particular white ball that was orinally in the second urn being chosen is $\frac{1}{24}$. There are 8 such white balls, so the probability of one of them being chosen is $\frac{8}{24}.$ The probability of a white ball being chosen is $\frac{21 \times 7}{24 \times 26}+\frac{8}{24}.$
I am about to give a detailed explanation in another way, When we are picking 7 balls at a time from First Urn (Say Urn A), it is trivial that, we could transfer a minimum of 2 white balls and a maximum of 7 white balls to the second urn (Say Urn B).
So, on the whole, the number of white balls picked from Urn A ranges from i = 2 to 7, and the picture below, explains how we calculate probability of picking white ball from Urn B, for each case.
Click here to view the image for case-by-case analysis
Since, probability (Column IV) in each case, has a weighted dependency on Probability of picking white ball in A (Column II). We multiply Column II and Column IV values for each case, and we sum up all those products, to get the Total probability of obtaining a white ball from B.
Mathematically, that summation can be written as,
$\displaystyle\sum\limits_{i=2}^7 {\frac{\left(\begin{array}{c}21\\ i\end{array}\right) \times \left(\begin{array}{c}5\\ (7-i)\end{array}\right) \times {(i + 8)}} {\left(\begin{array}{c}26\\ 7\end{array}\right) \times {24}}} $
Computing the above summation, would lead us to the probability around 0.5689.
If it seems strange, (or) anything I could clarify in my answer, kindly let me know.
Hope this helps.