Three bags each contain balls numbered from 1 to 10

Question

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.

Answer 1

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 $$\textrm{P}(x_{1}\leq{}k,x_{2}\leq{}k,x_{3}\leq{}k)=\textrm{P}(x_{1}\leq{}k)\textrm{P}(x_{2}\leq{}k)\textrm{P}(x_{3}\leq{}k)=\boxed{\Big(\frac{k}{10}\Big)^{3}}.$$

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:

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

Extended response after your edit:

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 $k-1$."