Exercise 12.2 - Chapter 12 - Introduction to Probability Theory - 11th Maths Guide Samacheer Kalvi Solutions

You can Download the Exercise 12.2 - Chapter 12 - Introduction to Probability Theory - 11th Maths Guide Samacheer Kalvi Solutions with expert answers for all chapters. Perfect for Tamil & English Medium students to revise the syllabus and score more marks in board exams. Download and share it with your friends

Share this to Friend on WhatsApp

Ex 12.2
Question 1.
If $A$ and $B$ are mutually exclusive events $P(A)=\frac{3}{8}$ and $P(B)=\frac{1}{8}$, then find
(i) $\mathrm{P}(\overline{\mathrm{A}})$
(ii) $P(A \cup B)$
(iii) $\mathrm{P}(\overline{\mathrm{A}} \cap \mathrm{B}$
(iv) $\mathrm{P}(\overline{\mathrm{A}} \cup \overline{\mathrm{B}})$
Solution:
(i)
$
\begin{aligned}
& \mathrm{P}(\mathrm{A})=\frac{3}{8}, \mathrm{P}(\mathrm{B})=\frac{1}{8} \\
& \mathrm{P}(\overline{\mathrm{A}})=1-\mathrm{P}(\mathrm{A})=1-\frac{3}{8}=\frac{5}{8}
\end{aligned}
$

(ii)
$
\begin{aligned}
\mathrm{P}(\mathrm{A} \cup \mathrm{B}) & =\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B}) \quad(\because \mathrm{A} \text { and } \mathrm{B} \text { are mutually exclusive }) \\
& =\frac{3}{8}+\frac{1}{8}=\frac{4}{8}=\frac{1}{2}
\end{aligned}
$

(iii)
$
\begin{aligned}
\mathrm{P}(\mathrm{A} \cap \mathrm{B}) & =\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cup \mathrm{B} \\
& =3 / 8+1 / 8-1 / 2
\end{aligned}
$

(iv)
$
\begin{aligned}
\mathrm{P}(\overline{\mathrm{A}} \cap \mathrm{B}) & =\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})=1 / 8-0=1 / 8 \\
\mathrm{P}(\overline{\mathrm{A}} \cup \overline{\mathrm{B}}) & =\mathrm{P}\left[(\mathrm{A} \cap \mathrm{B})^{\prime}\right]=1-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\
& =1-0=1
\end{aligned}
$
Question 2.
If $A$ and $B$ are two events associated with a random experiment for which $P(A)=0.35, P(A$ or $B)=0.85$, and $\mathrm{P}(\mathrm{A}$ and $\mathrm{B})=0.15$.
Find (i) $\mathrm{P}($ only B)
(ii) $\mathrm{P}(\overline{\mathrm{B}})$
(iii) $\mathrm{P}($ only A)

Solution:
$
\begin{aligned}
& \text { Given } P(A)=0.35 \\
& P(A \cup B)=0.85 \\
& P(A \cap B)=0.15
\end{aligned}
$
We know $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$
\begin{aligned}
& \text { (i.e.,) } 0.85=0.35+\mathrm{P}(\mathrm{B})-0.15 \\
& \Rightarrow 0.85-0.2=\mathrm{P}(\mathrm{B}) \\
& \text { (i.e.,) } \mathrm{P}(\mathrm{B})=0.65
\end{aligned}
$

(i) $\mathrm{P}($ only $\mathrm{B})=\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$=0.65-0.15=0.50$
(ii) $\mathrm{P}(\overline{\mathrm{B}})=1-\mathrm{P}(\mathrm{B})=1-0.65=0.35$
(iii) $\mathrm{P}(\mathrm{A}$ only $)=\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.35-0.15=0.20$
Question 3.
A die is thrown twice. Let Abe the event, 'First die shows 5' and B be the event, 'second die shows 5'. Find $P(A \cup B)$.
Solution:
When a die is throw twice $\mathrm{n}(\mathrm{s})=6^2=36$
Let $A$ be the event that first die shows 5 and $B$ be the event that second die shows 5 Now $A=\{(5,1),(5$,
2) $(5,3),(5,4),(5,5)(5,6\}$
$\mathrm{n}(\mathrm{A})=6 \Rightarrow \mathrm{P}(\mathrm{A})=\frac{n(\mathrm{~A})}{n(\mathrm{~S})}=\frac{6}{36}$
and $B=\{(1,5),(2,5),(3,5),(4,5),(5,5),(6,5)\}$
$\mathrm{n}(\mathrm{B})=6 \Rightarrow \mathrm{P}(\mathrm{B})=\frac{n(\mathrm{~B})}{n(\mathrm{~S})}=\frac{6}{36}$
Also $A \cap B=\{(5,5)\} \Rightarrow \mathrm{n}(\mathrm{A} \cap \mathrm{B})=1$
$\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$=\frac{6}{36}+\frac{6}{36}-\frac{1}{36}=\frac{11}{36}$
Question 4.
The probability of an event $A$ occurring is 0.5 and $B$ occurring is 0.3 . If $A$ and $B$ are mutually exclusive events, then find the probability of
(i) $\mathrm{P}(\mathrm{A} \cup \mathrm{B})$
(ii) $\mathrm{P}(\mathrm{A} \cap \bar{B})$
(iii) $\mathrm{P}(\bar{A} \cap \mathrm{B})$
Solution:
$\mathrm{P}(\mathrm{A})=0.5, \mathrm{P}(\mathrm{B})=0.3$

Here $A$ and $B$ are mutually exclusive.
(i) $P(A \cup B)=P(A)+P(B)$ $=0.5+0.3=0.8$
(ii) $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cup \mathrm{B})$ $=0.5+0.3-0.8$
$\mathrm{P}(\mathrm{A} \cap \mathrm{B})=0$
$\mathrm{P}(\mathrm{A} \cap \bar{B})=\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.5-0=0.5$
(iii) $\mathrm{P}(\bar{A} \cap \mathrm{B})=\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.3-0=0.3$
Question 5.
A town has 2 fire engines operating independently. The probability that a fire engine is available when needed is 0.96 .
(i) What is the probability that a fire engine is available when needed?
(ii) What is the probability that neither is available when needed?

Solution:
(i) $P$ (atleast one engine is available $)=(1-$ probability of no engine available $)$
$
\begin{aligned}
& =1-\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)=1-\mathrm{P}\left(\mathrm{A}^{\prime}\right) \mathrm{P}\left(\mathrm{B}^{\prime}\right) \\
& =1-(0.04)(0.04)=1-0.0016=0.9984 \\
& \text { (ii) } \mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)=\mathrm{P}\left(\mathrm{A}^{\prime}\right) \mathrm{P}\left(\mathrm{B}^{\prime}\right) \\
& =0.04 \times 0.04 \\
& =0.0016
\end{aligned}
$
Question 6.
The probability that a new railway bridge will get an award for its design is 0.48 , the probability that it will get an award for the efficient use of materials is 0.36 , and that it will get both awards is 0.2 . What is the probability, that (i) it will get atleast one of the two awards 00 it will get only one of the awards.
Solution:
Given $\mathrm{P}(\mathrm{A})=0.48, \mathrm{P}(\mathrm{B})=0.36$ and $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.2$

$\begin{aligned}
& \text { (i) } \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\
& =0.48+0.36-0.2=0.64 \\
& \text { (ii) } \mathrm{P}(\mathrm{Getting} \text { only one award }) \\
& =\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\
& =(0.48-0.2)+(0.36-0.2) \\
& =0.28+0.16=0.44
\end{aligned}$