mathematics-class-11-unit-06-chapter-03-probability-l-3-6-vwimnfkd35e

### <Success>Recall</Success>
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<ul>
<li>$P(A)$, where $A$ is an event corresponding to a random experment $E$ whose sample space is $\Omega$, is $\frac{\#A}{\#\Omega}.$</li>
<li>$P(A \cup B) = P(A) + P(B)$ if $A$ and $B$ are disjoint.</li>
<li>But if $A$ & $B$ are not disjoint then<br>
$P(A \cup B) = P(A) + P(B) - P(A \cap B).$</li>
<li>If $A$ & $B$ are independent events then<br>
$P(A \cap B) = P(A) \times P(B).$</li>
</ul>
</div>
<div style='float: left;text-align:left; width:1%;font-size:30px;'>
<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-03-probability-l-3-6-vwImnfkd35e-3.jpg"></p>
<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-03-probability-l-3-6-vwImnfkd35e-3a.jpg"></p>
</div>

### <Success>Solution</Success>
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<p>Let $X$ be the number of steps he takes to the right direction & $Y$ be the number of steps he takes to the left direction.</p>
<p>$\therefore X + Y = 6 \ \ \& \ |X - Y| = 2$</p>
</div>
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### <Success>Solution</Success>
<div style='float: left;text-align:left; width:100%;font-size:28px;'>
$\therefore$ We get   
<table>
<thead>
<tr>
<th>(i)</th>
<th>(ii)</th>
</tr>
</thead>
<tbody>
<tr>
<td>$X + Y = 6$</td>
<td>$X + Y = 6$</td>
</tr>
<tr>
<td>$X - Y = 2$</td>
<td>Y - X = 2$</td>
</tr>
<tr>
<td>$2X = 8$</td>
<td>$2Y = 8$</td>
</tr>
<tr>
<td>$\Rightarrow X = 4$</td>
<td>$\Rightarrow Y = 4$</td>
</tr>
<tr>
<td>$\Rightarrow Y = 2$</td>
<td>$\Rightarrow X = 2$</td>
</tr>
<tr>
<td>$(0.5)^6$</td>
<td>$(0.5)^6$</td>
</tr>
</tbody>
</table>
<p>$\therefore$ Desired Probability $ = 2(0.5)^6$</p>
<p>$\quad \quad \quad=(0.5)^5$</p>
</div>
<div style='float: left;text-align:left; width:1%;font-size:30px;'>
<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-03-probability-l-3-6-vwImnfkd35e-7.jpg"></p>

### <Success>Solution</Success>
<div style='float: left; width:100%;font-size:30px;'>
<p>Let $P(A) = x $ $\ \ \therefore P(B) = P(C) = x$</p>
<p>$\therefore P(A \cap B)=\frac{P(A)}{2}=\frac{x}{2}=P(B \cap C)=P(A \cap C)$</p>
<p>and $P(A \cap B \cap C)=\frac{x}{4}$<br>
$\therefore P(A \cup B \cup C) = P(A)+P(B)+P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C)$</p>
<p>$\Rightarrow 1 =3x+\frac{x}{4}- 3\frac{x}{2}$</p>
</div>
<p>======</p>
<h3 id="successsolutionsuccess"><Success>Solution</Success></h3>
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<p>$\Rightarrow 4 = 12x + x - 6x = 7x$</p>
<p>$\therefore x=\frac{4}{7}$<br>
$\therefore P(A \cap B \cap C^C) = P(A \cap B) - P(A \cap B \cap C)$</p>
<p>$=\frac{x}{2}-\frac{x}{4} =\frac{2}{7}-\frac{1}{7}$</p>
<p>$=\frac{1}{7}$</p>
</div>
<div style='float: left;text-align:left; width:1%;font-size:30px;'>
<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-03-probability-l-3-6-vwImnfkd35e-9.jpg"></p>

### <Success>Solution</Success>
<div style='float: left; width:100%;font-size:30px;'>
<p>A wins 2 matches. This can be done in ${}^{4}C_2$ ways</p>
<p>$W W L L$</p>
<p>$W L W L$</p>
<p>$W L L W$</p>
<p>$L W W L$</p>
<p>$L W L W$</p>
<p>$L L W W$</p>
<p>Then wins $5^{\text{th}}$ match $= W$</p>
</div>
<p>======</p>
<h3 id="successsolutionsuccess-1"><Success>Solution</Success></h3>
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<p>$\therefore$ The probability is :</p>
<p>$6\times(0.4)^2 \times(0.6)^2 \cdot(0.4)$</p>
<p>$(0.4)^3 \times6^2 \cdot(0.4)$</p>
<p>$\therefore$ P(A wins the tournament)</p>
<p>$=(0.4)^3+3(0.4)^3 \cdot 0.6+6 \cdot (0.6)^2 \cdot(0.4)^3$</p>
<p>$=(0.4)^3(1+1.8+2.16)$</p>
<p>$\approx 0.317 < 0.352$</p>
<p>A should choose best of 3 option.</p>
</div>
<div style='float: left;text-align:left; width:1%;font-size:30px;'>
<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-03-probability-l-3-6-vwImnfkd35e-17.jpg"></p>

### <Success>Solution</Success>
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<p>a) A is called before B.</p>
<ul>
<li>
<p>A is called at no 1. $\ \ $  A _ _ _ _</p>
<p>$\therefore$ Number of possibility $= 4!$</p>
</li>
<li>
<p>A is called at no 2.  $\ \ $ _ A _ _ _</p>
<p>$\therefore$ Number of possibility $= 3×3!$</p>
</li>
<li>
<p>A is called at no 3. $\ \ $ _ _ A _ _</p>
<p>$\therefore 3×2×2 = 12 $</p>
</li>
<li>
<p>A is called at no 4.  $\ \ $ _ _ _ A _</p>
</li>
</ul>
<p>$\therefore 3! = 6$</p>
</div>
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### <Success>Solution</Success>
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<p>$\therefore$ Number of possibility $\ = 24 + 18 + 12 + 6 = 60$</p>
<p>Now total number of parmutations of $5$ number is $5! = 120$<br>
$\therefore \ P(A \ \text{is called before} \ B) \ = \frac{60}{120}=\frac{1}{2}$</p>
</div>
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<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-03-probability-l-3-6-vwImnfkd35e-19.jpg">