# Three bags each contain balls numbered from 1 to 10

Three bags marked A, B and C each contain a set of balls numbered from 1 to 10. We pull a ball from each bag. Describe the appropriate sample space.Assuming all outcomes are equally likely calculate the probability of the events:

(i)$$A_k$$ = {the numbers on all three balls are less than or equal to k}, k = 1, 2, . . . , 10;

(ii) $$B_k$$ = {k is the greatest number chosen}, k = 1, 2, . . . , 10.

PS If requested I will provide you the solution in my book and explain exactly what I do not understand!

(a) Let k ∈ {1, 2, . . . , 10} be fixed. Then

$$P(A_k)$$ =(k/10)^3 (Here I do not understand at all how they derived this)

b) Observe that Ak−1 ⊆ Ak (why?) for every k = 2, 3, . . . 10, and that

$$B_k$$ = $$A_k$$ \ $$A_(k−1)$$ (why so/how?)

Hence, for every k = 2, 3, . . . 10, P(Ak) = P(Ak−1) + P(Bk), (again how is this derived?)

which by part (a) implies that

P(Bk) = P(Ak) − P(Ak−1) = ( k/ 10)^3 − ( (k − 1) 10 )^3 , k = 2, 3, . . . 10.

Furthermore, P(B1) = P(A1) = ( 1/ 10)^3 .

Therefore, P(Bk) = P(Ak) − P(Ak−1) = ( k/ 10)^3 − ( (k − 1)/ 10 )^3 , k = 1, 2, 3, . . . 10.

• explaining exactly what you do not understand would be helpful – Henry Apr 8 at 17:20
• Yes, it is requested that you show what you haven´t understood. – callculus Apr 8 at 17:54

Since you have distinguished the bags (i.e. named them "A", "B", and "C"), it seems appropriate to view the sample space as a set $$\mathcal{S}:=\{(1,1,1),(2,1,1),\dots,(10,10,10)\}$$ of $$10^{3}=1000$$ ordered triples. Since drawing from separate bags is an independent process and since $$\textrm{P}(x\leq{}k)=\frac{k}{10}$$ for each individual bag, it is clear that $$\textrm{P}(x_{1}\leq{}k)\cap\textrm{P}(x_{2}\leq{}k)\cap\textrm{P}(x_{3}\leq{}k)=\boxed{\Big(\frac{k}{10}\Big)^{3}}$$.

Now for part (ii). It is not difficult to see that $$\mathit{B}_{k}\subset\mathit{A}_{k}$$ because, rather than encompassing all outcomes with each ball less than or equal to $$k$$, $$\mathit{B}_{k}$$ describes the subset of $$\mathit{A}_{k}$$ where at least one ball $$x_{i}$$ has value $$k$$. To determine this probability, we should focus on the ball that must be $$k$$. Notice that $$\mathit{B}_{k}$$ is partitioned into the following three sets:

1. Outcomes with one $$k$$-valued ball
2. Outcomes with two $$k$$-valued balls
3. Outcomes with three $$k$$-valued balls

Clearly, we want our solution to include each of these outcome classes. Having broken the problem into cases, it is not difficult to proceed:

1. $$\textrm{P}(\textrm{exactly one ball} = k)=\Big(\binom{3}{1}\frac{1}{10}\Big)\cdot\Big(\frac{k-1}{10}\Big)^{2}=\frac{3(k-1)^{2}}{1000}$$ because there are three bags to choose the $$k$$-valued ball from and the balls from the remaining bags must be at most $$k-1$$ (refer to previous part).
2. $$\textrm{P}(\textrm{exactly two balls}=k)=\Big(\binom{3}{2}\big(\frac{1}{10}\big)^{2}\Big)\Big(\frac{k-1}{10}\Big)=\frac{3(k-1)}{1000}$$ by similar reasoning as (1).
3. $$\textrm{P}(\textrm{exactly three balls}=k)=\binom{3}{3}\big(\frac{1}{10}\big)^{3}=\frac{1}{1000}$$

So the probability that an outcome $$\Omega$$ is an element of $$\mathit{B}_{k}$$ equals the sum of the above three items:

$$\textrm{P}(\Omega\in\mathit{B}_{k})=\frac{3(k-1)^{2}}{1000}+\frac{3(k-1)}{1000}+\frac{1}{1000}=\boxed{\frac{3k^{2}-3k+1}{1000}}$$

To see that $$\mathit{A}_{k-1}\subset\mathit{A}_{k}$$, just think about the outcomes that these events describe. In words, $$\mathit{A}_{k-1}$$ is the set of outcomes where each ball is valued less than or equal to $$k-1$$. So of course all of these outcomes will be contained in $$\mathit{A}_{k}$$, because $$\mathit{A}_{k}$$ describes the outcomes with balls valued less than or equal to $$k$$.

To see that $$\mathit{A}_{k}=\mathit{A}_{k-1}+\mathit{B}_{k}$$ (this is equivalent to $$\textrm{P}(\mathit{A}_{k})=\textrm{P}(\mathit{A}_{k-1})+\textrm{P}(\mathit{B}_{k})$$), think closely about the partitioning that I used in the solution above. To be precise, $$\mathit{A}_{k}$$ can be partitioned into the sets of outcomes:

1. Those with exactly one $$k$$ in them (the other balls being less than k)
2. Those with exactly two $$k$$s in them (the other ball being less than k)
3. Those with exactly three $$k$$s in them
4. Those with exactly zero $$k$$s in them (all three balls being $$k-1$$ or less)

Notice that items (1), (2), and (3) make up $$\mathit{B}_{k}$$ as I outlined above and that item (4) is $$\mathit{A}_{k-1}$$ because it describes "the set of outcomes where each ball is valued no greater than $$k-1$$."

• Thanks a lot for the detailed explanation! – The Poor Jew Apr 9 at 1:34
• Could you just explain on (ii) the P(exactly one/two/three ball equal to k) why you have 3 choose 1/2/3? – The Poor Jew Apr 9 at 1:56
• Think of it this way: Let me take, for example, the outcomes where exactly two balls have the value k. This means one of the balls is valued strictly less than k. It would be erroneous to claim that this probability is (1/10)^{2}*(k-1)/10 because this only considers one way of "distributing the k-valued balls" among the bags. In particular, the two k-valued balls could be from bags A & B, B & C, xor A & C. So you can see that there are three (three choose two) ways I could have a particular unordered outcome. – Sam van der Poel Apr 9 at 2:09
• Understood it thanks again – The Poor Jew Apr 9 at 2:11