A jar contains five blue balls and five red balls. You roll a fair die once. Next you randomly draw (without replacement) as many balls from the jar as the number of points you have rolled with the die.
(a) What is the probability that all of the balls drawn are blue?
(b) What is the probability that the number of points shown by the die is $r$ given that all of the balls drawn are blue?
For a), if $A=[$all balls are blues$]$ and $H_i=[$result of the roll of dice$]$, we have:
$\mathbb{P}(A)=\mathbb{P}(\bigcup_{i=1}^{6}(A\cap H_i))=\sum_{i=1}^{6}\mathbb{P}(A\cap H_i)=\sum_{i=1}^{6}\mathbb{P}(H_i)\mathbb{P}(A|H_i)=\frac{1}{6}\sum_{i=1}^{6}\frac{\binom{5}{i}}{\binom{10}{i}}$
that I know is equal to $\frac{5}{36}$. How can I solve that division? I only arrive to say that $\frac{\binom{5}{i}}{\binom{10}{i}}=\frac{(10-i)!}{(5-i)!}$…
For b), if it's true that $\mathbb{P}(A)=\frac{5}{36}$, we have $\mathbb{P}(B_r|A)=\frac{\mathbb{P}(B_r\cap A)}{\mathbb{P}(A)}=\frac{\mathbb{P}(B_r)\mathbb{P}(A|B_r)}{\mathbb{P}(A)}=\frac{\frac{1}{6}\cdot\frac{\binom{5}{r}\binom{10-5}{r-r}}{\binom{10}{r}}}{\frac{5}{36}}$.
So my question is: does it exist a property to calculate $\frac{\binom{a}{b}}{\binom{c}{d}}$ or not?
Thanks for any help.
Best Answer
$\frac{\binom{5}{i}}{\binom{10}{i}}=\frac{5!}{10!}\frac{(10-i)!}{(5-i)!} = \frac{5!5!}{10!}\frac{(10-i)!}{5!(5-i)!} = \frac {5!5!} {10!} \times \, ^{(10-i)}C_5$
So, P = $ \frac {1} {6} \times \frac {5!5!} {10!} \times \, {\sum \limits_{i=1}^5 {^{(10-i)}}C_5}$ ....(i)
Please note that $i=6$ has zero probability of all balls being blue.
Now ${\sum \limits_{i=1}^5 {^{(10-i)}}C_5}$ is equivalent to ${\sum \limits_{i=5}^9 {^{i}}C_5}$.
Now we know, $\sum \limits_{r=k}^n {^{r}C_k} = \, ^{(n+1)}C_{(k+1)}$. It is the coefficient of $x^k$ in $1 + (1+x) + (1+x)^2...+(1+x)^n$ and there are ways to prove it including combinatorics. I am not proving it here but just giving you the expression.
So (i) becomes $ \frac {1} {6} \times \frac {5!5!} {10!} \times \, ^{10}C_6 = \frac {1} {6} \times \frac {5!5!} {10!} \times \frac {10!} {6!4!} = \frac{5}{36}$.