Exercise 12.2 - Chapter 12 Introduction to Probability Theory 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

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 $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.2$

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