### <Success style="font-size:25px"> Probability L-6 </Success> ### <primary style="font-size:24px"> Probability <br> lecture-16 </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;text-align:left'> </div> <div style='float: right; width:1%;'> <!--  --> </main> <hr> <footer style = "font-size: 18px"> $\rightarrow$ $\rightarrow$ <span style = "color:blue"> Probability lecture-16 </span> $\rightarrow$ Binomial distribution and use of binomial theorem in counting $\rightarrow$ Concept of random variable </footer>

### <Success style="font-size:25px"> Probability L-6 </Success> ### <primary style="font-size:24px"> Binomial distribution and use of binomial theorem in counting </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;text-align:left'> - Tossing a coin 3 times \& noting the sequence of outcomes. - TTT, TTH, THT, HTT, HHT, HTH, THA, HHH - Consider 100 tosses: therefore $2^{100}$ many sequence. </div> <div style='float: right; width:1%;'> <!--  --> </div> </main> <hr> <footer style = "font-size: 18px"> $\rightarrow$ Probability lecture-16 $\rightarrow$ <span style = "color:blue"> Binomial distribution and use of binomial theorem in counting </span> $\rightarrow$ Concept of random variable $\rightarrow$ Concept of random variable </footer>

### <Success style="font-size:25px"> Probability L-6 </Success> ### <primary style="font-size:24px"> Concept of random variable </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;text-align:left'> - A random variable is a maping for $\Omega \rightarrow \mathbb{R}$. - Consider $X$ to be a random variable such that - $X(\omega)$= Number of H's in the sequence. - TTT $ \rightarrow 0$ - TTH THT HTT $ \rightarrow 1$ - HTH HHT THH $ \rightarrow 2$ - HHH $ \rightarrow 3$ </div> <div style='float: right; width:1%;'> <!--  --> </div> </main> <hr> <footer style = "font-size: 18px">Probability lecture-16 $\rightarrow$ Binomial distribution and use of binomial theorem in counting $\rightarrow$ <span style = "color:blue"> Concept of random variable </span> $\rightarrow$ Concept of random variable $\rightarrow$ Concept of random variable </footer>

### <Success style="font-size:25px"> Probability L-6 </Success> ### <primary style="font-size:24px"> Concept of random variable </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;text-align:left'> - Notice that x takes values $\\{0,1,2,3\\}$ when no. of tosses $=3$ - $\therefore$ If numbers of tosses is $n$ then corresponding r.v. $X$ will take values: $\\{0,1, \cdots n\\}$ - Note that: The probability of $i, \quad i \in\\{0, 1, \ldots n\\}$ is not same $\forall i$ . - $ x: ~ 0 \quad 1 \quad 2 \quad 3$ - $\quad\quad 1 \quad 3 \quad 3 \quad 1$ </div> </main> <hr> <footer style = "font-size: 18px">Binomial distribution and use of binomial theorem in counting $\rightarrow$ Concept of random variable $\rightarrow$ <span style = "color:blue"> Concept of random variable </span> $\rightarrow$ Concept of random variable $\rightarrow$ Concept of random variable </footer>

### <Success style="font-size:25px"> Probability L-6 </Success> ### <primary style="font-size:24px"> Concept of random variable </primary> <hr> <main> <div style='float: left; width:99%;font-size:25px;text-align:left'> - Suppose $p$ is the probability that the coin gives a H in one toss. - $ P(T T T)=(1-p)(1-p)(1-p)=(1-p)^3 = q^3 $ - $P(1 H)=P(T T H)+P(T H T)+P(H T T)$ - $=q q p +q p q + p q q = 3 p q^2 $ - Similarly, $P(2 H)=3 p^2 q$ - And, $ P(3 H)= p^3 $ - We can see that the probabilities can be obtained by - $ (q+p)^3 \quad \\& $ - $P(X=i) ={ }^3 C_i p^i(1-p)^{3-i} $ </div> </main> <hr> <footer style = "font-size: 18px">Concept of random variable $\rightarrow$ Concept of random variable $\rightarrow$ <span style = "color:blue"> Concept of random variable </span> $\rightarrow$ Concept of random variable $\rightarrow$ Problem-1 </footer>

### <Success style="font-size:25px"> Probability L-6 </Success> ### <primary style="font-size:24px"> Concept of random variable </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;text-align:left'> - In general if a coin has probability 'p' for $H$ then the r.v X : - giving the can not of $H$ in $n$ tosses can be obtained from $(q+p)^n$ and, - P(x=i)=$ {n_ {C_i}} p^i q^{n-i}$ - Probability mass function for Binomial distribution with parameter n \& p Bin (n, p) </div> </main> <hr> <footer style = "font-size: 18px">Concept of random variable $\rightarrow$ Concept of random variable $\rightarrow$ <span style = "color:blue"> Concept of random variable </span> $\rightarrow$ Problem-1 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Probability L-6 </Success> ### <primary style="font-size:24px"> Problem-1 </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;text-align:left'> - Suppose we have $X$ following Bin(6,p) - It is given that 9 P(4) = P(2) - what in the value of 'p'? - Solution: - $ P(4)={}^6 {C_ 4} p^4 q^2 \quad P(2)={ }^6 C_2 p^2 q^4 $ - $ \therefore \frac{P(2)}{P(4)}=9 $ - i.e $ \frac{{ }^6 C_ 2 p^2 q^4}{6_ {C_4} p^4 q^2}=9 $ </div> </main> <hr> <footer style = "font-size: 18px">Concept of random variable $\rightarrow$ Concept of random variable $\rightarrow$ <span style = "color:blue"> Problem-1 </span> $\rightarrow$ Solution $\rightarrow$ Problem-2 </footer>

### <Success style="font-size:25px"> Probability L-6 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;text-align:left'> - i.e $ \frac{q^2}{p^2} = 9 $ - i.e $ \frac{q}{p}=3 \therefore q= 3 p$ - Now $ p+q=1 ~ \text{and} ~ p +3 p=1 \Rightarrow p =0.25 $ or $ \frac{1}{4} $ </div> <div style='float: right; width:1%;'> <!--      --> </div> </main> <hr> <footer style = "font-size: 18px">Concept of random variable $\rightarrow$ Problem-1 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-2 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Probability L-6 </Success> ### <primary style="font-size:24px"> Problem-2 </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;text-align:left'> - Suppose $X$ follows Bin(n, p) when n = 8. - P(1)=0.2048 - P(2)=0.1024 - Find the value of p. - Solution: - $ 8_ {C_1} p^1 q^7 =0.2048$ - $8_ {C_2}^{p^2 q^6} =0.1024 $ - On dividing, </div> </main> <hr> <footer style = "font-size: 18px">Problem-1 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-2 </span> $\rightarrow$ Solution $\rightarrow$ Mean or expectation </footer>

### <Success style="font-size:25px"> Probability L-6 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;text-align:left'> - $ \frac{8 p q^7}{\frac{8! p^2 q^6}{2! 6!} } =\frac{0.2048}{0.1024} \text { or } \quad \frac{8 p q^7}{28 p^2 q^6}=2 $ - or = $\frac{2 q}{7 p} =2 \Rightarrow q=7 p$ - $\therefore$ value of $p=\frac{1}{8}$. </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Problem-2 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Mean or expectation $\rightarrow$ Problem-3 </footer>

### <Success style="font-size:25px"> Probability L-6 </Success> ### <primary style="font-size:24px"> Mean or expectation </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;text-align:left'> - An important concept of a r.v is its mean or expectation. - The mean of a Bin(n, p) r.v =n p, $ ~ $ and its variance is $ ~ n p q$ </div> </main> <hr> <footer style = "font-size: 18px">Problem-2 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Mean or expectation </span> $\rightarrow$ Problem-3 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Probability L-6 </Success> ### <primary style="font-size:24px"> Problem-3 </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;text-align:left'> - If the mean of a Bin(n, p) r.v is 4 and its variance in half of its mean. Then compute the probability that the r.v. takes value $\geq 2$. - Solution: - Mean =n p and variance = n p r \& given that - n p q= $\frac{1}{2} n p \Rightarrow q=\frac{1}{2} $ - $ \therefore p=\frac{1}{2} $ - Mean = 4 $\Rightarrow n \cdot \frac{1}{2}=4$ $\Rightarrow n=8 $. - $ \therefore Bin(8, \frac{1}{2}) $ </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Mean or expectation $\rightarrow$ <span style = "color:blue"> Problem-3 </span> $\rightarrow$ Solution $\rightarrow$ Generating function </footer>

### <Success style="font-size:25px"> Probability L-6 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;text-align:left'> - We need to compute $P\left(Bin \left(8, \frac{1}{2}\right)\right) \geq 2$ - ie $ 1-P(0)-P(1) $ - $= 1-{ }^8 C_ 0 (\frac{1}{2})^0(\frac{1}{2})^8-{ }^8 C_1\left(\frac{1}{2}\right)^1\left(\frac{1}{2}\right)^7 $ - = $ 1-\left(\frac{1}{2}\right)^8-8 \cdot\left(\frac{1}{2}\right)^8 $ - = $ 1-9 \cdot\left(\frac{1}{2}\right)^8 $ - = $ 1-9 \cdot \frac{1}{256}=\frac{256-9}{256} $ - = $ \frac{247}{256} $ </div> <div style='float: right; width:1%;'> <!--     --> </div> </main> <hr> <footer style = "font-size: 18px">Mean or expectation $\rightarrow$ Problem-3 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Generating function $\rightarrow$ Generating function </footer>

### <Success style="font-size:25px"> Probability L-6 </Success> ### <primary style="font-size:24px"> Generating function </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;text-align:left'> - Use of Binomial theorem in counting the number of certain events. - For illustration. Consider the problem that in how many ways you can have 5 different positive integers - $0<n_1<n_2<n_3<n_4<n_5$ - $\text {such that} \quad \sum_{i=1}^5 x_i=20$ - The number of solution in 7 . </div> </main> <hr> <footer style = "font-size: 18px">Problem-3 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Generating function </span> $\rightarrow$ Generating function $\rightarrow$ Problem-4 </footer>

### <Success style="font-size:25px"> Probability L-6 </Success> ### <primary style="font-size:24px"> Generating function </primary> <hr> <main> <div style='float: left; width:99%;font-size:15px;text-align:left'> - But we have done it in a mechanical way: - such a solution is not guaranteed to give all possible combination | 1 | 2 | 3 | 4 | 10 | | -------- | -------- | -------- | -------- | ---- | | 1 | 2 | 3 | 5 | 9 | | 1 | 2 | 3 | 6 | 8 | | 1 | 2 | 4 | 5 | 8 | | 1 | 2 | 4 | 6 | 7 | | 1 | 3 | 4 | 5 | 7 | | 2 | 3 | 4 | 5 | 6 | - The concept is "Generating Function" essentially a power series which may be finite or infinite. </div> <div style='float: right; width:1%;'> <!--   --> </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Generating function $\rightarrow$ <span style = "color:blue"> Generating function </span> $\rightarrow$ Problem-4 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Probability L-6 </Success> ### <primary style="font-size:24px"> Problem-4 </primary> <hr> <main> <div style='float: left; width:99%;font-size:15px;text-align:left'> - In how many ways we can have two variables - $x \ \\& \ y \ \Rightarrow \ 0 \leq x \leq 2 \quad$ \& $\quad 1 \leq y \leq 2$ - \& $ {x+y}=3 \text { here } X \text { and } Y \text { are integer variables }$ | $x$ | $y$ | sum | | -------- | -------- | -------- | | 0 | 1 | 1 | | 0 | 2 | 2 | | 1 | 1 | 2 | | 1 | 2 | 3 | | 2 | 1 | 3 | | 2 | 2 | 4 | </div> </main> <hr> <footer style = "font-size: 18px">Generating function $\rightarrow$ Generating function $\rightarrow$ <span style = "color:blue"> Problem-4 </span> $\rightarrow$ Solution $\rightarrow$ Example-1 </footer>

### <Success style="font-size:25px"> Probability L-6 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;text-align:left'> - Consider the following two polynomials \& compute their product. - $ \left(z^0+z^1+z^2\right)\left(z^1+z^2\right) $ - $= \left(1+z+z^2\right)\left(z+z^2\right) $ - $= z+z^2+z^3+z^2+z^3+z^4 $ - $= z+2 z^2+2 z^3+z^4 . $ </div> <div style='float: right; width:1%;'> <!--   --> </div> </main> <hr> <footer style = "font-size: 18px">Generating function $\rightarrow$ Problem-4 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Example-1 $\rightarrow$ Example-1 </footer>

### <Success style="font-size:25px"> Probability L-6 </Success> ### <primary style="font-size:24px"> Example-1 </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;text-align:left'> - Suppose we want to find out $x+y+z=10$ - when, $ ~ 0 \leq X \leq 4$, $Y>0$ - $Z \geq 0$ here $X, Y, Z$ are integer variables. - Consider, $ \left(x^0+x^1+x^2+x^3+x^4\right) $ - $ \cdot\left(x+x^2+\cdots\right)\left(1+x+x^2+\cdots\right) $ </div> </main> <hr> <footer style = "font-size: 18px">Problem-4 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Example-1 </span> $\rightarrow$ Example-1 $\rightarrow$ Example-1 </footer>

### <Success style="font-size:25px"> Probability L-6 </Success> ### <primary style="font-size:24px"> Example-1 </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;text-align:left'> - Obtain the coefficient of $x^{0}$ that will give the number of possible solution to the problem. - $\left(1+x+x^2+x^3+x^4\right) x $ - $ \cdot\left(1+x+x^2+\cdots\right)^2 $ - $\left(x+x^2+x^3+x^4+x^5\right)\left(1+x+x^2 \cdots \right)^2$ </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Example-1 $\rightarrow$ <span style = "color:blue"> Example-1 </span> $\rightarrow$ Example-1 $\rightarrow$ Example-1 </footer>

### <Success style="font-size:25px"> Probability L-6 </Success> ### <primary style="font-size:24px"> Example-1 </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;text-align:left'> - Now we know that, - $ \left(1+x+x^2+\cdots\right)^n=\frac{1}{(1-x)^n}=(1-x)^{-n} $ - $ =1+\sum_ {r=1}^{\infty}{ }^{n-1+r} C_r x^r $ - we need to find out coefficient of $x^9, x^8, x^7. . . x^5$ - Let us consider $x^5$ - Here n=2, r=5 - $\therefore $ coefficient of $x^5=^{2+5-1}C_5$ </div> </main> <hr> <footer style = "font-size: 18px">Example-1 $\rightarrow$ Example-1 $\rightarrow$ <span style = "color:blue"> Example-1 </span> $\rightarrow$ Example-1 $\rightarrow$ Example-1 </footer>

### <Success style="font-size:25px"> Probability L-6 </Success> ### <primary style="font-size:24px"> Example-1 </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;text-align:left'> - $r=5 = {}^6 C_5 =6$ - Let us compute for $x^9$. - Here, $r=9 \therefore$ coeff $x^9= ^{9+2-1} C_5$ - ${ }^{10} C_9=10 $ </div> </main> <hr> <footer style = "font-size: 18px">Example-1 $\rightarrow$ Example-1 $\rightarrow$ <span style = "color:blue"> Example-1 </span> $\rightarrow$ Example-1 $\rightarrow$ Example-2 </footer>

### <Success style="font-size:25px"> Probability L-6 </Success> ### <primary style="font-size:24px"> Example-1 </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;text-align:left'> - You understand that the total no. of - possible solution = 10+9+8+7+6 = 40 - $\therefore$ The number of possible solution is 40 . </div> <div style='float: right; width:1%;'> <!--      --> </div> </main> <hr> <footer style = "font-size: 18px">Example-1 $\rightarrow$ Example-1 $\rightarrow$ <span style = "color:blue"> Example-1 </span> $\rightarrow$ Example-2 $\rightarrow$ Example-2 </footer>

### <Success style="font-size:25px"> Probability L-6 </Success> ### <primary style="font-size:24px"> Example-2 </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;text-align:left'> - Note that we are forming the power series corresponding to each variable by considering the values that it may take. - For example: In how many ways we can get $x+y+z=50$ - $\rightarrow x $ in a multiple of 2 - $y$ is a multiple of 3 - $z$ is positive \& multiple of 5 </div> </main> <hr> <footer style = "font-size: 18px">Example-1 $\rightarrow$ Example-1 $\rightarrow$ <span style = "color:blue"> Example-2 </span> $\rightarrow$ Example-2 $\rightarrow$ Problem-5 </footer>

### <Success style="font-size:25px"> Probability L-6 </Success> ### <primary style="font-size:24px"> Example-2 </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;text-align:left'> - $\therefore$ The no. of solution for x + y + z = 50 with the above restiction will be given by coeff. of $x^{50}$ in: - $\left(1+x^2+x^4+\cdots\right) \times(1+x^3+x^6+x^9+\cdots)$ - $ \times\left(x^5+x^{10}+x^{15}+\cdots\right)$ </div> </main> <hr> <footer style = "font-size: 18px">Example-1 $\rightarrow$ Example-2 $\rightarrow$ <span style = "color:blue"> Example-2 </span> $\rightarrow$ Problem-5 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Probability L-6 </Success> ### <primary style="font-size:24px"> Problem-5 </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;text-align:left'> - $ 0<n_1<n_2<n_3<n_4<n_5 \quad \sum_{i=1}^5 n_i=20$ - Note that all the 5 variables are integers. - Solution: - let : $ m_2=n_2-n_1 $ - $ m_3=n_3-n_2 $ - $ m_4=n_4-n_3 $ - $ m_5=n_5-n_4 $ - $\therefore n_1 $\& $ m_2 m_3 m_n m_5$ are all $>0$ </div> </main> <hr> <footer style = "font-size: 18px">Example-2 $\rightarrow$ Example-2 $\rightarrow$ <span style = "color:blue"> Problem-5 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Probability L-6 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;text-align:left'> - More over - $ n_1+n_2+n_3+n_4+n_5=20 $ - 5 $ n_1+4\left(n_2-n_1\right)+3\left(n_3-n_2\right)+2\left(n_4-n_3\right)+\left(n_5-n_4\right) =20 $ - or $5 n_1+4 m_2+3 m_3+2 m_4+m_5=20$ when $n_1 2$ \& all $m_i>0$ </div> </main> <hr> <footer style = "font-size: 18px">Example-2 $\rightarrow$ Problem-5 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Probability L-6 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;text-align:left'> - <ins>Let us now make the following subsitution.</ins> - $ x_1=m_1-1 $ - $ x_2=m_2-1 $ - $ x_3=m_3-1 $ - $ x_4=m_4-1 $ - $ x_5=m_5-1 $ </div> </main> <hr> <footer style = "font-size: 18px">Problem-5 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Probability L-6 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;text-align:left'> - $ \therefore$ Each $x_i \geq 0 $ - $ \therefore {5 x_ 1+4 x_ 2+3 x_ 3+2 x_ 4+x_ 5}={5\left(x_ 1-1\right)+4\left(x_2-1\right)} $ - $ +3\left(m_3-1\right) $ $ +2\left(m_n-1\right) $ $ \text { and }\left(m_5-1\right) $ - $ =20-(1+2+3+4+5) $ $ =20-15=5 $ when $ x_i \geq 0 $ </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Probability L-6 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;text-align:left'> - Now $5 x_1$, take values {0, 5 , 10.........} - $4 x_2$ take value{0,4,8,12........... } - $3 x_3$ take value{0,3,6,........... } - $2 x_4$ take value{0,2,4,6........... } - $x_5$ take value{0,1,2,3........... } - $\therefore$ The power series for us is - $ (1+x^5+x^{10}\cdots)(1+x^4+x^8+\cdots) (1+x^3+x^6+\cdots) $ - $ \left(1+x^2+x^4+\cdots\right) \left(1+x+x^2+x^3-\cdots\right) $ </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Probability L-6 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;text-align:left'> - We need to compute the coeff. of $x^5$. - $\therefore$ The problem in to compute coeff of $x^5$ in. - $ \left(1+x^5\right)\left(1+x^4\right)\left(1+x^3\right)\left(1+x^2+x^4\right)\left(1+x+x^2+x^3+x^4+x^5\right) $ - $ \left(1+x^4+x^5\right)\left(1+x^2+x^4+x^3+x^5+x^7\right)\left(1+x+x^2+x^3+x^4+x^5\right) $ </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Exercise </footer>

### <Success style="font-size:25px"> Probability L-6 </Success> ### <primary style="font-size:24px"> Solution </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;text-align:left'> - $ \left(1+x^4+x^5\right)\left(1+x^2+x^3+x^4+x^5\right)\left(1+x+x^2+x^3+x^4+x^5\right) $ - $\left(1+x^2+x^3+x^4+x^5+x^4+x^5\right)\left(1+x+\cdots+x^5\right) $ - $ =\left(1+x^2+x^3+2 x^4+2 x^5\right)\left(1+x+\cdots+x^5\right) $ - $ \therefore$ coeff. $x^5=7$ - No of possible solution =7 </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Exercise $\rightarrow$ Thank you </footer>

### <Success style="font-size:25px"> Probability L-6 </Success> ### <primary style="font-size:24px"> Exercise </primary> <hr> <main> <div style='float: left; width:99%;font-size:30px;text-align:left'> - $ 0<n_1<n_2<n_3<n_4 $ - $\rightarrow \sum_{i=1}^n n_i=16$ Note that all the 4 variables are integers. - The no. of solution = 9 </div> <div style='float: right; width:1%;'> <!--        --> </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Exercise </span> $\rightarrow$ Thank you $\rightarrow$ </footer>

### <Success style="font-size:25px"> Probability L-6 </Success> ### <primary style="font-size:24px"> Thank you </primary> <hr> <main> <div style='float: right; width:1%;'> <!--  --> </div> </main> <hr> <footer style = "font-size: 18px">Solution $\rightarrow$ Exercise $\rightarrow$ <span style = "color:blue"> Thank you </span> $\rightarrow$ $\rightarrow$ </footer>