Exercise 12.-Additional Questions - Chapter 12 - Introduction to Probability Theory - 11th Maths Guide Samacheer Kalvi Solutions

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Additional Problems
Question 1.
$A$ and $B$ are two events associated with random experiment for which $P(A)=0.36, P(A$ or $B)=0.90$ and $P(A$ and $B)=0.25$. Find
(i) $\mathrm{P}(\mathrm{B})$
(ii) $\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})$
Solution:
(i) Given $\mathrm{P}(\mathrm{A})=0.36, \mathrm{P}(\mathrm{A} \cup \mathrm{B})=0.09, \mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.25$
$P(A \cup B)=P(A)+P(B)-P(A \cap B)$
(i.e.,) $0.90=0.36+P(B)-0.25$
$0.90=0.11+\mathrm{P}(\mathrm{B})$
$\therefore \mathrm{P}(\mathrm{B})=0.90-0.11=0.79$
(ii) $\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})=\mathrm{P}\left\{\left(\mathrm{A}^{\prime} \cup \mathrm{B}\right)^{\prime}\right\}$ (Demorgan Law)
$P(A \cup B)^{\prime}=1-P(A \cup B)=1-0.90=0.1$.
Question 2 .
Given $\mathrm{P}(\mathrm{A})=0.5, \mathrm{P}(\mathrm{B})=0.6$ and $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.24$. Find
(i) $P(A \cup B)$
(ii) $P(\overline{\mathrm{A}} \cap \mathrm{B})$
(iii) $P(A \cap \bar{B})$
(iv) $\mathrm{P}(\overline{\mathrm{A}} \cup \overline{\mathrm{B}})$
(v) $\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})$
Solution:
(i) $\mathrm{P}(\mathrm{A})=0.5, \mathrm{P}(\mathrm{B})=0.6, \mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.24$
$\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
(i.e.) $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=0.5+0.6-0.24$
$=1.1-0.24=0.86$
$\therefore \mathrm{P}(\mathrm{A} \cup \mathrm{B})=0.86$
(ii) $\mathrm{P}(\overline{\mathrm{A}} \cap \mathrm{B})=\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$=0.6-0.24=0.36$
(iii) $\mathrm{P}(\mathrm{A} \cap \overline{\mathrm{B}})=\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$=0.5-0.24=0.26$
(iv) $\mathrm{P}(\overline{\mathrm{A}} \cup \overline{\mathrm{B}})=\mathrm{P}\left\{(\mathrm{A} \cap \mathrm{B})^{\prime}\right\}=1-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$=1-0.24=0.76$
(v) $\left.\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})=\mathrm{P}\{\mathrm{A} \cup \mathrm{B})^{\prime}\right\}=1-\mathrm{P}(\mathrm{A} \cup \mathrm{B})$
$=1-0.86=0.14$.

Question 3.
The probability of an event $A$ occurring is 0.5 and $\mathrm{B}$ occurring is 0.3 . If $\mathrm{A}$ and $\mathrm{B}$ are mutually exclusive events, then find the probability of neither A nor B occurring.
Solution:
Given $A$ and $B$ are mutually exclusive and $P(A)=0.5, P(B)=0.3$
$
\therefore \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})=0.5+0.3=0.8
$
So, $\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right)=\mathrm{P}\left\{(\mathrm{A} \cup \mathrm{B})^{\prime}\right\}=1-\mathrm{P}(\mathrm{A} \cup \mathrm{B})$
$
=1-0.8=0.2
$
Question 4.
The probability that a new ship will get an award for its design is 0.25 , the probability that it will get an award for the efficient use of materials is 0.35 and that it will get both awards is 0.15 . What is the probability, that (/) it will get atleast one of the two awards (ii) it will get only one of the awards?
Solution:
Probability of getting the award for its design $=P(A)=0.25$
Probability of getting the award for the efficient use of materials $=P(B)=0.35$
Probability of getting both awards $=P(A \cap B)=0.15$
Now $\mathrm{P}(\mathrm{A})=0.25 ; \mathrm{P}(\mathrm{B})=0.35$ and $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.15$
$\therefore(\mathrm{i}) \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$=0.25+0.35-0.15=0.60-0.15=0.45$
(ii) $\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}^{\prime}\right.$ or $\left.\mathrm{B} \cap \mathrm{A}^{\prime}\right)=\mathrm{P}\left(\mathrm{A} \cap \mathrm{B}^{\prime}\right)+\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}\right)$
$\mathrm{P}\left(\mathrm{A} \cap \mathrm{B}^{\prime}\right)=\mathrm{P}(\mathrm{A})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$=0.25-0.15=0.10$
$\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}\right)=\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$
$=0.35-0.15=0.20$
$\therefore \mathrm{P}\left(\mathrm{A} \cap \mathrm{B}^{\prime}\right)+\mathrm{P}\left(\mathrm{A}^{\prime} \cap \mathrm{B}\right)=0.10+0.20=0.30$.