<h3 id="successprobability-lecture-5success"><Success>Probability lecture-5</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>Problem 1</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>Bag containing 12 R 28 G balls. Three balls are drawn in succession without replacement.</p> <p>We know that $P$ ($1^{st}$ ball drawn is Red)</p> <p>$P\left(R_1\right)=\frac{12}{20}=\frac{3}{5} \ \ \& \ \ P\left(G_1\right)=\frac{8}{20}=\frac{2}{5}$<br> We have calculated some conditional probabilities</p> <p>say $P\left(G_2 \mid R_1\right)=\frac{8}{19}$</p> <p>What is the unconditional probability that the $2^{n d}$ ball drawn in Green? $\text { i.e } P\left(G_2\right)$</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-05-probability-l-5-6-8j7wy3ibfvI-2.jpg">

### <Success>Solution</Success> <div style='float: left;text-align:left; width:60%;font-size:30px;'> <p>Now the event $G_2=R_1 G_2 \quad G_1 G_2$</p> <p>$P\left(G_2\right)=P\left(R_1 G_2 \cup G_1 G_2\right)$</p> <p>$\therefore P\left(G_2\right) =P\left(R_1 \cap G_2\right)+P\left(G_1 \cap G_2\right) $</p> <p>$=P\left(G_2 \mid R_1\right) \cdot P\left(R_1\right)+P\left(G_2 \mid G_1\right) \cdot P\left(G_1\right) $</p> <p>$=\frac{8}{19} \times \frac{3}{5}+\frac{7}{19} \times \frac{2}{5}=\frac{24+14}{19 \times 5}\\=\frac{38}{19 \times 5}=\frac{2}{5}$</p> </div> <div style='float: right; width:40%;'> <!----> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/a3db296a-3a1f-40af-9f2e-db2521ef4200/public"></p>

### <Success>Unconditional probability</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>What is the unconditional probability that the $3^{\text {rd }}$ ball drawn in Red?</p> <p><strong>Solution:</strong></p> <p>The $3^{\text {rd }}$ ball can be Red in union of four disjoint events:</p> <p>$R_1 \cap R_2 \cap R_3 \cup R_1 \cap G_2 \cap R_3 \cup G_1 \cap R_2 \cap R_3\cup G_1 \cap G_2 \cap R_3 $</p> <p>$\therefore P\left(R_3\right)=P\left(R_1 R_2 R_3\right)+P\left(R_1 G_2 R_3\right) +P\left(G_1 R_2 R_3\right) $</p> <p>$+P\left(G_1 G_2 R_3\right)$</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-05-probability-l-5-6-8j7wy3ibfvI-5.jpg"></p>

### <Success>Solution</Success> <div style='float: left;text-align:left; width:50%;font-size:30px;'> <p>$P\left(R_3\right)=\frac{3}{5} \times \frac{11}{19} \times \frac{10}{18}+\frac{3}{5} \times \frac{8}{19} \times \frac{11}{18}+\frac{2}{5} \times \frac{12}{19} \times \frac{11}{18}+\frac{2}{5} \times \frac{7}{19} \times \frac{12}{18} $</p> <p>$=\frac{1}{5 \times 19 \times 18}(330+264+264+168)\\ =\frac{1026}{5 \times 19 \times 18}=\frac{3}{5}$</p> </div> <div style='float: right; width:40%;'> <!----> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/dc5d7d62-f5d1-4ec9-0caa-3fb534a21600/public"></p>

### <Success>Problem 2</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>Suppose you have 3 fair dice $\ni P(1)=P(2) =P(3)=\cdots=P(6)=\frac{1}{6}$</p> <p>Also you have a fake dice which has four faces as $5$ & two faces as $6$</p> <p>You choose 1 of the four dice at random and throw it. You get a 5</p> <p>(a) What is the probability of getting a 5 ?</p> <p>(b) Given that you have got a 5 what is the probability that you have chosen the fake die?</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-05-probability-l-5-6-8j7wy3ibfvI-7.jpg"></p>

### <Success>Solution</Success> <div style='float: left;text-align:left; width:50%;font-size:30px;'> <p>(a) $P(5)=P(5 / \text { fair die }) \times P(\text { fair die }) $</p> <p>$+P(5 / \text { fake die }) \times P(\text { fake die}) $</p> <p>$=\frac{1}{6} \times \frac{3}{4}+\frac{2}{3} \times \frac{1}{4} $</p> <p>$=\frac{3}{24}+\frac{4}{24}=\frac{7}{24} $</p> <p>(b) $P(\text { chosen the fake die } \mid \text { result is 5) } \longrightarrow$ bayes’ theorem</p> </div> <div style='float: right; width:50%;'> <!----> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/8939a23f-dcec-4714-4615-4537a38c6400/public"></p>

### <Success>Bayes' theorem</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>$\quad P(A \mid B)=\frac{P(A \cap B)}{P(B)}=\frac{P(B \mid A) \cdot P(A)}{P(B)}$</p> <p>$P(\text { fake die } \mid 5) =\frac{P(5 \cap \text { fake die })}{P(5)} $</p> <p>$=\frac{P(5 \mid \text { fake die }) \times P(\text { fake die })}{P(5)} $</p> <p>$=\frac{\frac{2}{3} \times \frac{1}{4}}{\frac{7}{24}}=\frac{2}{12} \times \frac{24}{7}=\frac{4}{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-05-probability-l-5-6-8j7wy3ibfvI-9.jpg">

### <Success>Problem 3</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>Suppose you have 10 coins numbered $1,2,3 \ldots 10$. Suppose probability of getting a $H$ in single tose of the $i^{\text {th }}$ coin in $\frac{i}{10}, i=1,2, \ldots 10$.</p> <p>You choose a coin randomly and tose it</p> <p>If the result you got in a $H$, what is the probability 5 that yon have chosen the $5^{th}$ coin?</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-05-probability-l-5-6-8j7wy3ibfvI-10.jpg">

### <Success>Solution</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>Let $E$ be the event that you choose the $5^{th}$ coin & $B$ be the event that you got a $H$ we want to compute $P(E \mid B)$. Using Bayes’ Theorem $$ P(E \mid B)=\frac{P(B \cap E)}{P(B)}=\frac{P(B \mid E) \times P(E)}{P(B)} $$</p> <p>Now $P(B \mid E)=\frac{5}{10}=\frac{1}{2}$. $2 P(E)=\frac{1}{10}$ as the coin is chosen randmoly.</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-05-probability-l-5-6-8j7wy3ibfvI-11.jpg"></p>

### <Success>Solution</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>$\therefore P(E \mid B)=\frac{\frac{1}{2} \times \frac{1}{10}}{P(B)}=P(\text { gating a } H) $</p> <p>Now</p> <p>$P(B)=P\left(1^{st} coin \cap B\right)+P\left(2^d \cap B\right) +\cdots+P\left(10^{\text {th }} \cap B\right)$</p> <p>$=P(H \mid 1^{st} coin ). P(1^{st} coin) + \dots + P(H \mid 10^{th} coin). P(10^{th} coin)$</p> <p>$=\frac{1}{10} \times \frac{1}{10}+\frac{2}{10} \times \frac{1}{10}+\cdots+\frac{10}{10} \times \frac{1}{10} $</p> <p>$=\frac{1}{100}(1+2+\cdots+10)=\frac{1}{100} \times \frac{10 \times 11}{2}=\frac{11}{20} $</p> <p>$\therefore P(E \mid B) =\frac{\frac{1}{20}}{\frac{11}{20}}=\frac{1}{11}$</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-05-probability-l-5-6-8j7wy3ibfvI-12.jpg"></p>

### <Success>Problem 4</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>Suppose a student is answering an $M C Q$ question with 5 options of which only one is correct. Now let $p$ be the probability that the student guesses the answer i.e he ticks randmoly & let $1-p$ be the probability that he knows the answer and therefore ticks correctly.</p> <p>Suppose the student ticked correctly. What is the probability that he guessed the answer.</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-05-probability-l-5-6-8j7wy3ibfvI-13.jpg"></p>

### <Success>Solution</Success> <div style='float: left;text-align:left; width:100%;font-size:28px;'> <p>We want to compute $P$ (guessing\ticked righting)</p> <p>Let $E$ be the event that he guessed it & $B$ be the event that he ticked it correctly</p> <p>$\therefore P(E \mid B)=\frac{P(B \mid E) \cdot P(E)}{P(B)}$</p> <p>Now $P(B \mid E)=\frac{1}{5},$ & $P(E)=p$ (given)</p> <p>And $P(B) =P(B \mid \text {guess}) \cdot P(\text {guess})+P(B \mid \text {know}) \cdot P(\text {know}) $</p> <p>$=\frac{1}{5} \times p+1 \cdot(1-p)=\frac{p+5(1-p)}{5}=\frac{5-4 P}{5} $</p> <p>$\therefore P(E \mid B) =\frac{p}{5}$ \ $\frac{5-4 P}{5}=\frac{p}{5-4 P}$</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-05-probability-l-5-6-8j7wy3ibfvI-14.jpg"></p>

### <Success>Problem 5</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>Suppose you have 3 bags $A, B$ & $C$. The contents are as follows:</p> <p>$A: 1W+2G+3 R $</p> <p>$B: 2W+1 G+1 R $</p> <p>$C: 4W+5G+3 R .$</p> <p>You choose a bag at random & take out two balls from it. Suppose you get $1W$ & $1R$ ball. What is the probability that you have choose Bag A.?</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-05-probability-l-5-6-8j7wy3ibfvI-15.jpg"></p>

### <Success>Solution</Success> <div style='float: left;text-align:left; width:100%;font-size:25px;'> <p>Let $E$ be the event of selecting bag $A$. <br> $B$ be the event of getting $1W+1R$.</p> <p>We want to compute $P(E \mid B)$</p> <p>Using Bayers’ Theorem: $P(E \mid B)=\frac{P(B \mid E) \cdot P(E)}{P(B)}$</p> <p>Now $P(E)=\frac{1}{3}$ and $P(B \mid E)=\frac{3}{{}^6{c_2}}=\frac{3}{\frac{6!}{2!4!}}$</p> <p>$=\frac{3}{\frac{5 \times 6}{2}}=\frac{1}{5} .$</p> <p>Similary ,$P(1W+1 R \mid Bag B)^2=\frac{2}{{}^4c_2}=\frac{2}{\frac{4 !}{2 ! 2 !}}=\frac{2}{6}=\frac{1}{3} $</p> <p>$P(1W+1 R \mid Bag C)=\frac{12}{12 c_2}=\frac{12}{\frac{11\times12}{2}}=\frac{2}{11}$</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-05-probability-l-5-6-8j7wy3ibfvI-16.jpg">

### <Success>Solution</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>$\therefore P(B) =P(1W+1R) $</p> <p>$=\frac{1}{5} \times \frac{1}{3}{}+\frac{1}{3} \times \frac{1}{3}+\frac{1}{3} \times \frac{2}{11} $</p> <p>$=\frac{1}{3}\left(\frac{1}{5}+\frac{1}{3}+\frac{2}{11}\right)=\frac{1}{3}\left(\frac{33+55+30}{165}\right) $</p> <p>$=\frac{1}{3} \times \frac{118}{165} $</p> <p>$\therefore P(E \mid B) =\frac{\frac{1}{5} \times \frac{1}{3}}{\frac{1}{3} \times \frac{118}{165}}=\frac{1}{5} \times \frac{165}{118}=\frac{33}{118}$</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-05-probability-l-5-6-8j7wy3ibfvI-17.jpg">

### <Success>Problem 6</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>Suppose you are waiting for a letter from KOLKATA, KATHMANDU and MUSCAT.</p> <p>Suppose you receive a letter and the only thing legible in the address is two successive letters: AT</p> <p>What is the probability that it is from KOLKATA?</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-05-probability-l-5-6-8j7wy3ibfvI-18.jpg">

### <Success>Solution</Success> <div style='float: left;text-align:left; width:100%;font-size:25px;'> <p>$P\left(K O L K A T A \mid AT\right) =\frac{P\left(AT \mid \text { KOLKATA }\right) \times P(K O L K A T A)}{P\left(AT\right)}$</p> <p>Now</p> <p>$P(A T \mid \text { KOLKATA) } : \frac{1}{6}$</p> <p>$P(A T \mid \text { KATHMANDU)} : \frac{1}{8} $</p> <p>$P(A T \mid \text { MUSCAT) }=\frac{1}{5}$</p> <p>$P(K O L K A T A \mid A T) =\frac{\frac{1}{6} \times \frac{1}{3}}{\frac{1}{6} \times \frac{1}{3}+\frac{1}{8} \times \frac{1}{3}+\frac{1}{5} \times \frac{1}{3}}$</p> <p>$=\frac{\frac{1}{6}}{\frac{1}{6}+\frac{1}{8}+\frac{1}{5}} =\frac{\frac{1}{6}}{\frac{40+30+48}{5 \times 6 \times 8}} $</p> <p>$=\frac{40}{40+30+48}=\frac{40}{118}=\frac{20}{59}$</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-05-probability-l-5-6-8j7wy3ibfvI-19.jpg"></p>

### <Success>Problem 7</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>Suppose India & Australia are playing a five -match test series. Suppose Virat Kohli won the toss in the first four matches.</p> <p>What is the probability that he will win the toss in the fifth match as well?<br> (A) 1</p> <p>(B) $\frac{1}{5}$</p> <p>(B) 5</p> <p>(C) 0</p> <p>(D) $\frac{1}{2}$</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-05-probability-l-5-6-8j7wy3ibfvI-20.jpg"></p>

### <Success>Solution</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>We assume that the toss is done using an unbiased coin, so that in each toss the probability of winning is equal for both the captain.</p> <p>Can it be 1? Because since Virat won all the previous tosses, therefore he will win the fifth toss.</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-05-probability-l-5-6-8j7wy3ibfvI-21.jpg">

### <Success>Solution</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>Can it be $\frac{1}{5}$ ?</p> <p>Because Virat won all the 4 toss therefore now the Australia captain will have $\frac{4}{5}$ probability of winning.</p> <p>Can it be 0 ?</p> <p>Because now Australia captain will have to win tho toss.</p> <p>$\therefore P$ (Virat winning) $=0$</p> <p>Correct option is (D). ie Answer is $\frac{1}{2}$ $P$ (Virat winning $5^{\text {th }}$ toss $\mid$ he won all the previous tosses) in same as $P$ (Virat winning the toss) as these are independent event.</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-05-probability-l-5-6-8j7wy3ibfvI-22.jpg"></p> </div> <p>======</p> <h3 id="successthank-yousuccess"><Success>Thank you</Success></h3> <div style='float: left;text-align:left; width:100%;font-size:30px;'>