<h3 id="successprobability-lecture-2success"><Success>Probability lecture-2</Success></h3> <div style='float: left; width:50%;'> </div> <div style='float: right; width:0%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-06-chapter-01-probability-l-1-6-bsayy-ke2ok-2.jpg"></p>

### <Success>Recall the formula</Success> <div style='float: left;text-align:left; width:99%;font-size:30px;'> <ol> <li>If we have $n$ identical balls which are to be kept in $k$ boxes such that no box will remain empty then the number of possible arrangement is $ {}^{n-1}C_{k-1}$</li> </ol> <p>Ex: If we have 3 identical balls to be kept in 2 boxes such that no box is empty then there are two possible way of doing it : $(1,2)$ or $(2,1)$.</p> </div> <div style='float: right; width:1%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-06-chapter-02-probability-l-2-6-ibhu4weckbm-3.jpg"> <img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-06-chapter-02-probability-l-2-6-ibhu4weckbm-3a.jpg"></p>

### <Success>Recall the formula</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <ol start="2"> <li>$n$ identical balls to be kept in $k$ boxes such that some boxes may remain empty then the number of possible arrangements is<br> ${}^{n+k-1}{C_{k-1}}.$</li> </ol> <p><ins>Ex</ins>. 3 balls 2 boxes</p> <p>$\therefore$ possible arrangements are.</p> <p>$$(0,3) \quad(1,2) \quad(2,1) \quad(3,0)=4$$</p> <p>and we have ${}^{3+2-1}C_{2-1}={ }^4 c_1=4$</p> </div> <div style='float: right; width:1%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-06-chapter-02-probability-l-2-6-ibhu4weckbm-4.jpg"></p>

### <Success>Application</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>In how many ways you can choose 5 numbers $a, b, c, d, e $ such that each one is $>0$ and</p> <p>$a+b+c+d+e=20$</p> <p>then the number of possible solution, is</p> <p>$ { }^{20-1}c_{5-1}={ }^{19} c_4 $</p> <p>However, if $a, b, c, d, e \geq 0$ then the number of possible solution is ${}^{20+5-1}C_{5-1}$ $={ }^{24} C_4$</p> </div> <div style='float: right; width:1%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-06-chapter-02-probability-l-2-6-ibhu4weckbm-5.jpg"></p>

### <Success>Problem 1</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>In how many ways you can choose 5 numbers $n_1, n_2, n_3, n_4$ & $n_5$ such that $\quad n_i>0 \quad \forall i=1,5$ and $\quad{n_1<n_2<n_3<n_4<n_5}$</p> <p>$$ \sum_{i=1}^5 n_i=20 $$ $\therefore$ Note that $:(1,2,3,4,10)$ is a possible solution.</p> <p>But $(1,2,4,4,9)$ in not a solution since 4 in repeated.</p> </div> <div style='float: right; width:1%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-06-chapter-02-probability-l-2-6-ibhu4weckbm-7.jpg"></p>

### <Success>Solution</Success> <div style='float: left; width:100%;font-size:28px;'> <p>Note that smallest possible value for $n_1=1$ for $n_2=2$ & for $n_5=5$</p> <p>$\therefore$ Let us define 5 new variables $x_1 x_2, x_3, x_n$ & $x_5$ as follows:</p> <p>$ x_1=n_1-1$</p> <p>$ x_2=n_2-2$</p> <p>$ x_3=n_3-3$</p> <p>$ x_4=n_4-4$</p> <p>$ x_5=n_5-5$</p> <p>$ \therefore \quad \text { Each } \ x_i \ \text{is} \geq 0 \quad x_1 \leq x_2 \leq x_3 \leq x_4 \leq x_5 $</p> </div> <div style='float: right; width:1%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-06-chapter-02-probability-l-2-6-ibhu4weckbm-8.jpg"></p>

### <Success>Solution</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>Hence the problem becomes: to choose 5 numbers,</p> <p>$x_1$ & $x_2$ & $x_3$ & $x_n$ & $x_5 \ $ such that $\text { all are } \geq 0$</p> <p>$ x_1+x_2+x_3+x_4+x_5$</p> <p>$=n_1-1+n_2-2+n_3-3+n_4-4 $</p> <p>$=\sum_{i=1}^5 n_i-(1+2+3+4+5)$</p> <p>$ =20-15 =5 .$</p> </div> <div style='float: right; width:1%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-06-chapter-02-probability-l-2-6-ibhu4weckbm-9.jpg"></p>

### <Success>Solution</Success> <div style='float: left;text-align:left; width:50%;font-size:30px;'> <table> <thead> <tr> <th>$x_1$</th> <th>$x_2$</th> <th>$x_3$</th> <th>$x_4$</th> <th>$x_5$</th> <th></th> <th></th> <th></th> <th></th> <th></th> <th></th> </tr> </thead> <tbody> <tr> <td>0</td> <td>0</td> <td>0</td> <td>0</td> <td>5</td> <td>:</td> <td>1</td> <td>2</td> <td>3</td> <td>4</td> <td>10</td> </tr> <tr> <td>0</td> <td>0</td> <td>0</td> <td>1</td> <td>4</td> <td>:</td> <td>1</td> <td>2</td> <td>3</td> <td>5</td> <td>9</td> </tr> <tr> <td>0</td> <td>0</td> <td>0</td> <td>2</td> <td>3</td> <td>:</td> <td>1</td> <td>2</td> <td>3</td> <td>6</td> <td>8</td> </tr> <tr> <td>0</td> <td>0</td> <td>1</td> <td>1</td> <td>3</td> <td>:</td> <td>1</td> <td>2</td> <td>4</td> <td>5</td> <td>8</td> </tr> <tr> <td>0</td> <td>0</td> <td>1</td> <td>2</td> <td>2</td> <td>:</td> <td>1</td> <td>2</td> <td>4</td> <td>6</td> <td>7</td> </tr> <tr> <td>0</td> <td>1</td> <td>1</td> <td>1</td> <td>2</td> <td>:</td> <td>1</td> <td>3</td> <td>4</td> <td>5</td> <td>7</td> </tr> <tr> <td>1</td> <td>1</td> <td>1</td> <td>1</td> <td>1</td> <td>:</td> <td>2</td> <td>3</td> <td>4</td> <td>6</td> <td>6</td> </tr> </tbody> </table> <p>$\therefore \ \text{number of possible solution is} \ 7 $</p> </div> <div style='float: right; width:1%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-06-chapter-02-probability-l-2-6-ibhu4weckbm-10.jpg"></p>

### <Success>Problem 2</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>In how many ways you choose 4 numbers</p> <p>${n_1<n_2<n_3<n_4}$</p> <p>$ {\sum_{i=1}^n n_i=16}, \quad n_1>0 \quad \forall i=1,2,3,4$</p> </div> <p>======</p> <h3 id="successsolutionsuccess"><Success>Solution</Success></h3> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>we define: $x_1=n_1-1 \\ x_2=n_2-2 \\ x_3=n_3-3 \\ x_n=n_4-4$</p> <p>such that each $\quad$ $x_i \geq 0 $</p> <p>$ x_1 \leqslant x_2 \leqslant x_3 \leqslant x_4$</p> <p>$\sum x_i=16-10=6$</p> </div> <div style='float: right; width:1%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-06-chapter-02-probability-l-2-6-ibhu4weckbm-11.jpg"></p>

##### <Success>Solution</Success> <div style='float: left;text-align:left; width:100%;font-size:28px;'> <table> <thead> <tr> <th>$x_1$</th> <th>$x_2$</th> <th>$x_3$</th> <th>$x_4$</th> </tr> </thead> <tbody> <tr> <td>0</td> <td>0</td> <td>0</td> <td>6</td> </tr> <tr> <td>0</td> <td>0</td> <td>1</td> <td>5</td> </tr> <tr> <td>0</td> <td>0</td> <td>2</td> <td>4</td> </tr> <tr> <td>0</td> <td>0</td> <td>3</td> <td>3</td> </tr> <tr> <td>0</td> <td>1</td> <td>1</td> <td>4</td> </tr> <tr> <td>0</td> <td>1</td> <td>2</td> <td>3</td> </tr> <tr> <td>1</td> <td>1</td> <td>1</td> <td>3</td> </tr> <tr> <td>1</td> <td>1</td> <td>2</td> <td>2</td> </tr> <tr> <td>0</td> <td>2</td> <td>2</td> <td>2</td> </tr> </tbody> </table> <p>$9$ Possible solution</p> </div> <div style='float: right; width:1%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-06-chapter-02-probability-l-2-6-ibhu4weckbm-12.jpg">

### <Success>Probability</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>We talk about probability when the underlying experiment is random.</p> <p>$\therefore$ the sample space $\Omega$ is known.</p> <p>And we need to compute probability of what??</p> </div> <div style='float: right; width:1%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-06-chapter-02-probability-l-2-6-ibhu4weckbm-13.jpg"></p>

### <Success>Example</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <ul> <li> <p>Coin tossing 5 times.</p> <ul> <li>what is the probability of two heads?</li> <li>what is the probability of the number of tales is prime?</li> </ul> </li> <li> <p>Passenger with bags:</p> <ul> <li>What is the probability that number of bags is even number of passenger is odd?</li> </ul> </li> </ul> </div> <div style='float: right; width:1%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-06-chapter-02-probability-l-2-6-ibhu4weckbm-14.jpg"></p>

### <Success>Event</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>An event is a subset of the sample space $\Omega$.</p> <p>consider : we toss a coin 3 times.</p> <p>we want to compute that number of heads is odd. $\mathrm{HHH}, \mathrm{HTT}, \mathrm{THT}, \mathrm{TTH}$.</p> <p>We are looking at the probability of the subset: $\{H T T, TH T, T T H, H H H\}$.</p> </div> <div style='float: right; width:1%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-06-chapter-02-probability-l-2-6-ibhu4weckbm-15.jpg">

### <Success>Definitions</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>The subsets of cardinality 1 are called <strong>elementary events</strong>.</p> <p>i.e The outcome of one individual trial in called an elementary event.</p> <p>If we throw a die, then number of elementary events is 6 $$ \{1\}, \{2\}, \{3\}, \cdots \{6\}. $$</p> <p>An event comprising of more that one elementary event is called a <strong>compound event</strong>.</p> </div> <div style='float: right; width:1%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-06-chapter-02-probability-l-2-6-ibhu4weckbm-16.jpg"></p>

### <Success>Example</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>Suppose the $\Omega$ is $\{1,2,3,4,5,6,7,8,9\}.$<br> How many compound events are possible?</p> <p><strong>Solution:</strong></p> <p>Since $|\Omega|=9$</p> <p>$\therefore$ there are $2^9$ possible subsets of which 1 is $\phi$ & 9 are elementary events.</p> <p>Number of compound events is</p> <p>$2^9-(9+1)=512-10=502$</p> </div> <div style='float: right; width:1%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-06-chapter-02-probability-l-2-6-ibhu4weckbm-17.jpg">

### <Success>Definitions</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <ul> <li> <p>Two events $E_1$ & $E_2$ are said to be <strong>disjoint</strong> if $E_1 \cap E_2=\phi$</p> <ul> <li>Eg: Throwing a die<br> $E_1$ : getting a number $\leq 3$.<br> $E_2$ : getting a number $> 4$.</li> <li>$E_1 \ \& \ E _2$ are disjoint.</li> </ul> </li> <li> <p>A sequence of events $E_1, E_2 \cdots E_K$ is said to be <strong>mutually exclusive</strong> if $E_i \cap E_j=\phi \ \ \text{for all} \ i,j \in \{1,2, \cdots k\} \ \ i\neq j$.</p> </li> </ul> </div> <div style='float: right; width:1%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-06-chapter-02-probability-l-2-6-ibhu4weckbm-18.jpg"></p>

### <Success>Independent events</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>Two events $A$ and $B$ are said to be independent if</p> <p>$P(A \cap B)=P(A) \times P(B)$.</p> </div> <div style='float: right; width:1%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-06-chapter-02-probability-l-2-6-ibhu4weckbm-19.jpg"></p>

##### <Success>Probability of an event</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>Probability is a mapping from the power set of $\Omega$ to $[0,1]$<br> i.e. if $A \subseteq \Omega$ then $P(A)=p | 0\leqslant p \leqslant 1$</p> <p>Where $P$ satisfies the following:<br> <strong>a)</strong> $P(A) \geq 0 \quad \forall A \subseteq \Omega$.<br> <strong>b)</strong> $P(\Omega)=1$<br> <strong>c)</strong> If $A_1, A_2 \cdots A_k$ are matually exculsive then<br> $\quad P\left(A_1 \cup A_2 \cdots \cup A_K\right)=\sum_{i=1}^k P\left(A_i\right)$</p> <p>Given a set $A \subseteq\Omega $</p> <p>$ P(A)$ is computed as, $\ \frac{\text {Number of elements in A}}{|\Omega|}$</p> <p>when the outcomes are equally likely.</p> </div> <div style='float: right; width:1%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-06-chapter-02-probability-l-2-6-ibhu4weckbm-20.jpg"></p>

### <Success>Example</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p><strong>1)</strong> Throwing a die and probability of getting an even number is</p> <p>$\frac{\# \{2,4,6\}}{\mid\Omega\mid}$</p> <p>$=\frac{3}{6}=\frac{1}{2} \text {. }$</p> <p><strong>2)</strong> Suppose probability of getting a H is $p$. when $0 < p < 1$.</p> <p>Then what is the probability of getting $2 \mathrm{H’s}$ in 3 tosses?</p> </div> <div style='float: right; width:1%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-06-chapter-02-probability-l-2-6-ibhu4weckbm-21.jpg"></p>

##### <Success>$\text{Property:} \ P\left(A^c\right)=1-P(A)$</Success> <div style='float: left;text-align:left; width:100%;font-size:28px;'> <p>Show that: $P\left(A^c\right)=1-P(A)$.</p> <p><strong>Proof:</strong></p> <p>Since $A \cup A^C=\Omega$</p> <p>$1= P(\Omega)=P\left(A \cup A^c\right)=P(A)+P\left(A^c\right) \\ P\left(A^c\right)=1-P(A) .$</p> <p>When we are tossing a coin,</p> <p>if $A=\{H\}$ then $A^c=\{T\}$</p> <p>$ \therefore$ if $P(H)=P$ then</p> <p>$P(T)=1-P \text {. }$</p> </div> <div style='float: right; width:1%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-06-chapter-02-probability-l-2-6-ibhu4weckbm-22.jpg"></p>

### <Success>Example</Success> <div style='float: left;text-align:left; width:100%;font-size:25px;'> <p>Suppose probability of getting a H is $p$. when $0 < p < 1$. Then what is the probability of getting $2 \mathrm{H’s}$ in 3 tosses?</p> <p><strong>Solution:</strong></p> <p>Now consider getting $2 \mathrm{H}$ ’s in 3 tosses, $ P(H H T)+P(H T H)+P(T H H) $</p> <ul> <li>$P(H H T)$ <ul> <li>$1^{st} H \rightarrow p$</li> <li>$2^{nd} H \rightarrow p$</li> <li>$3^{rd} T \rightarrow (1-p)$</li> </ul> </li> <li>$P(H T H) \ \rightarrow p^2(1-p)$</li> <li>$P(T H H) \ \rightarrow p^2(1-p)$</li> </ul> <p>Total probability $\ = 3p^2(1-p)$</p> </div> <div style='float: right; width:1%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-06-chapter-02-probability-l-2-6-ibhu4weckbm-23.jpg"></p>

### <Success>Property</Success> <div style='float: left;text-align:left; width:70%;font-size:30px;'> <p>Another important property is</p> <p>$P(A \cup B)=P(A)+P(B)-P(A \cap B) \text {. }$<br> $P(A \cup B)= P\left(B \cap A^c\right)+P(B \cap A)+P(A \cap B^C)$<br> $\because$ These three are disjoint.</p> </div> <div style='float: right; width:30%;'> <!-- --> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/ab3650aa-c11c-42e4-a697-ab807fdc0e00/public"></p> </div> <p>======</p> <h3 id="successpropertysuccess"><Success>Property</Success></h3> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>Now $P(B \cap A^c)=P(B)- P(A \cap B)$<br> $\& \ P(B \cap A)+P(A \cap B C)=P(A)$</p> <p>$\therefore$ Together we have , $P(A \cup B)=P(A)+P(B)-P(A \cap B)$</p> </div> <!--<div style='float: right; width:40%;'>--> <!-- --> <!---->

### <Success>Problem 3</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>Suppose $E$ is a random experiment. Let $A$ and $B$ be two events such that</p> <p>$0<P(A)<1$ & $0<P(B)<1$</p> <p>Then which of the following statements are true?</p> <p>(A) $A$ and $A^c$ are mutually exclusive.</p> <p>(B) $A$ and $A^c$ are independent.</p> <p>(C) $A$ and $B$ are independent $\Rightarrow$ $A$ and $B^c$ are independent.</p> <p>(D) $A$ and $B$ are independent $\Rightarrow$ $A^c$ and $B^c$ are independent.</p> </div> <div style='float: right; width:1%;'> <p><img alt="image" src="https://temp-public-img-folder.s3.ap-south-1.amazonaws.com/sathee.prutor.images/subject-images/iitpal/image/mathematics-class-11-unit-06-chapter-02-probability-l-2-6-ibhu4weckbm-25.jpg"></p>

### <Success>Solution</Success> <div style='float: left;text-align:left; width:70%;font-size:30px;'> <p>(A) It is obvious that $A$ & $A^c$ are matually exclusive.</p> <p>$\because\quad \nexists \ \omega \ \ \omega\in A \quad$& $\quad\omega \in A^C$</p> <p>(B) $A$ and $A^c$ are independent.<br> Now we know that $A$ and $B$ are independent if $P(A \cap B)=P(A) P(B)$</p> </div> <div style='float: right; width:30%;'> <!----> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/70ac0554-2ef5-4d26-71ce-4ab63d824a00/public"></p> </div> <p>======</p> <h3 id="successsolutionsuccess-1"><Success>Solution</Success></h3> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>Now $P\left(A \cap A^c\right)=0 $</p> <p>$ \because \quad A \cap A^c$ is $\phi$</p> <p>$P(A \cap A^c) = \frac{\# \phi}{\# \Omega}=0$</p> <p>However, $P(A).P\left(A^C\right) \neq 0 $<br> $\because 0 < P(A) < 1.$<br> $\therefore $ B is false.</p> </div> <!--<div style='float: right; width:40%;'>--> <!----> <!---->

### <Success>Solution</Success> <div style='float: left;text-align:left; width:70%;font-size:30px;'> <p>(C) $A$ & $B$ are independent $\Rightarrow A < B^c$ are independent.<br> $P\left(A \cap B^C\right) =P(A)-P(A \cap B)$</p> <p>$ =P(A)-P(A) P(B)$<br> $\{\because \ A \ \& \ B \ \text{are independent.}\}$</p> <p>$ P\left(A \cap B^C\right) =P(A)(1-P(B)) $</p> <p>$ =P(A) P(B^C)$</p> <p>$\therefore A\ \& \ B^C \text { are independent }$<br> $\therefore $ C is true.</p> </div> <div style='float: right; width:30%;'> <!----> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/a6310256-4063-440b-f47c-e4c8f67b1e00/public"></p>

### <Success>Solution</Success> <div style='float: left;text-align:left; width:70%;font-size:30px;'> <p>(D) $A$ and $B$ are independent $\Rightarrow$ $A^c$ and $B^c$ are independent</p> <p>$ P\left(A^c \cap B^c\right) =P(\Omega)-P(A \cup B)$$ $$=1-(P(A)+P(B)-P(A \cap B)) $</p> <p>$ =1-P(A)-P(B)+P(A \cap B) $</p> <p>$=(1-P(A))-P(B)(1-P(A)) $<br> $ \{ \because \ P(A \cap B)= P(A).P(B)\}$<br> $=(1-P(A))(1-P(B))$<br> $=P\left(A^c\right) \cdot P\left(B^c\right)$<br> $\therefore (A^c)$ & $(B^c)$ are independent.</p> </div> <div style='float: right; width:30%;'> <!----> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/b2b6e207-2302-45a7-f92e-aeda75068700/public"></p> </div> <p>======</p> <h3 id="successthank-yousuccess"><Success>Thank you</Success></h3> <div style='float: left;text-align:left; width:50%;font-size:30px;'> </div>