Probability of real life situations

I will be solving this question by calculating total number of outcomes and number of favourable outcomes and dividing them to get the answer.

Total number of ways is clearly $11^{6}$ , since there are total $11$ ways to select marks in each section of $A$ as well as $B$ .

$Calculating\quad total\quad number\quad of\quad favourable\quad outcomes$

We firstly consider number of ways marks of maths can be seleted that is common to both $A$ and $B$ and that is $=11$

Since their total marks are same hence we can conclude that the sum of physics and chemistry section of both $A$ and $B$ are same.

Now we will be counting the number of outcomes of selecting physics and chemistry

Sum of physics and chemistry section for both $A$ and $B$ can vary from $0$ to $20$

For the sum I will take two cases.

Case-1

Consider the sum to be $k$ where , $0 \le k \le10$ .

Then total number of ways for $A$ is equal to :

No. of non-negative integral solution of ${x}_{1}+{x}_{2}=k$ , that is equal to $k+1$

And same number of ways exist for B and hence total number of ways $= (k+1)^{2}$

Vary $k$ from $0$ to $10$ and total number of ways $\displaystyle {S}_{1}=\sum _{ k=0 }^{ 10 }{ { (k+1) }^{ 2 } }=\sum _{ k=1 }^{ 11 }{ { k }^{ 2 } }$

Case-2

For any $k$ , $11 \le k \le 20$ , we have for any case total number of ways for $A$ as :

No. of solutions of ${x}_{1}+{x}_{2}=k$ where $0 \le {x}_{1},{x}_{2} \le 10$

and that is equal to $21-k$ , that you can check for yourself.

And same number of ways exist for $B$ hence the total ways is ${(21-k)}^{2}$

Vary $k$ from $11$ to $20$ and you get the total number of ways as $=\displaystyle \sum _{ k=11 }^{ 20 }{ { (21-k) }^{ 2 } } =\sum _{ k=1 }^{ 10 }{ { k }^{ 2 } }$

Finally total number of ways of selecting physics and chemistry is given by :

$=\displaystyle \sum _{ k=1 }^{ 11 }{ { k }^{ 2 } } +\sum _{ k=1 }^{ 10 }{ { k }^{ 2 } }=(11) \times (81)$

So total number of favourable outcomes is equal to :

$={11}^{2} \times 81$

And our probability is $=\dfrac { 81 }{ { 11 }^{ 4 } } ={ \left(\dfrac { 3 }{ 11 } \right) }^{ 4 }$

$a=3 , b=11 , c=4 , a+b+c=18$