Exercise 12.3-Additional Questions - Chapter 12 Introduction to Probability Theory 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Additional Problems
Question 1.

If $\mathrm{P}(\mathrm{A})=0.4, \mathrm{P}(\mathrm{B})=0.7$ and $\mathrm{P}(\mathrm{B} / \mathrm{A})=0.5$, find $\mathrm{P}(\mathrm{A} / \mathrm{B})$ and $\mathrm{P}(\mathrm{A} \cup \mathrm{B})$. Solution:
$
\begin{aligned}
& \mathrm{P}(\mathrm{B} / \mathrm{A})=0.5 \Rightarrow \frac{\mathrm{P}(\mathrm{B} \cap \mathrm{A})}{\mathrm{P}(\mathrm{A})}=0.5 \\
& \text { (i.e.,) } \frac{\mathrm{P}(\mathrm{B} \cap \mathrm{A})}{0.4}=0.5 \\
& \therefore \mathrm{P}(\mathrm{B} \cap \mathrm{A})=0.4 \times 0.5=0.2 \\
& \text { (i.e.) } \mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.2 \\
& \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})
\end{aligned}
$

$
\begin{aligned}
& \mathrm{P}(\mathrm{A} \cup \mathrm{B})=0.4+0.7-0.2=0.9 \\
& \mathrm{P}(\mathrm{A} / \mathrm{B})=\frac{P(A \cap B)}{P(B)}=\frac{0.2}{0.7}=\frac{2}{7}
\end{aligned}
$
Question 2.
If $\mathrm{A}$ and $\mathrm{B}$ are two events such that $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{5}{6}, \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{3}, \mathrm{P}(\bar{B})=\frac{1}{2}$ show that $\mathrm{A}$ and $\mathrm{B}$ are independent.
Solution:
$
\begin{aligned}
& \mathrm{P}(\bar{B})=\frac{1}{2} \text { (i.e.,) } 1-\mathrm{P}(\mathrm{B})=\frac{1}{2} \\
& \therefore \quad 1-\frac{1}{2}=\mathrm{P}(\mathrm{B}) \\
& \text { (i.e.,) } \quad \mathrm{P}(\mathrm{B})=\frac{1}{2} \\
& P(A \cup B)=P(A)+P(B)-P(A \cap B) \\
& \text { (i.e.,) } \quad \frac{5}{6}=\mathrm{P}(\mathrm{A})+\frac{1}{2}-\frac{1}{3} \\
& \therefore \quad \frac{5}{6}-\frac{1}{2}+\frac{1}{3}=\mathrm{P}(\mathrm{A}) \\
& \text { (i.e.,) } \\
& P(A)=\frac{5-3+2}{6}=\frac{4}{6}=\frac{2}{3} \\
& \text { Now } \mathrm{P}(\mathrm{A})=\frac{2}{3}, \mathrm{P}(\mathrm{B})=\frac{1}{2} \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{3} \\
&
\end{aligned}
$
$\therefore \mathrm{A}$ and $\mathrm{B}$ are independent
Question 3.
$P(A)=0.3, P(B)=0.6$ and $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.25$. Find
(i) $P(A \cup B)$
(ii) $P(A / B)$
(iii) $P(B / \bar{A})$
(iv) $\mathrm{P}(\overline{\mathrm{A}} / \mathrm{B})$
(v) $\mathrm{P}(\overline{\mathrm{A}} / \overline{\mathrm{B}})$
Solution:

(i)
$
\begin{aligned}
\mathrm{P}(\mathrm{A} \cup \mathrm{B}) & =\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B}) \\
& =0.3+0.6-0.25 \\
& =0.9-0.25=0.65=\frac{65}{100}=\frac{13}{20}
\end{aligned}
$

(ii)
$
P(A / B)=\frac{P(A \cap B)}{P(B)}=\frac{0.25}{0.6}=\frac{25}{60}=\frac{5}{12}
$

(iii)
$
\begin{aligned}
\mathrm{P}(\mathrm{B} / \overline{\mathrm{A}}) & =\frac{\mathrm{P}(\mathrm{B} \cap \mathrm{A})}{\mathrm{P}(\overline{\mathrm{A}})} \\
\therefore \quad \mathrm{P}(\mathrm{B} \cap \overline{\mathrm{A}}) & =\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})=0.6-0.25 \\
& =0.35 \\
\therefore \quad \mathrm{P}(\overline{\mathrm{A}}) & =1-\mathrm{P}(\mathrm{A})=1-0.3=0.7 \\
\therefore \quad \mathrm{P}(\mathrm{B} / \overline{\mathrm{A}}) & =\frac{0.35}{0.7}=\frac{1}{2}
\end{aligned}
$

(iv)
$
\begin{aligned}
\mathrm{P}(\overline{\mathrm{A}} / \mathrm{B}) & =\mathrm{P}(\overline{\mathrm{A}} \cap \mathrm{B}) / \mathrm{P}(\mathrm{B}) \\
\mathrm{P}(\overline{\mathrm{A}} \cap \mathrm{B}) & =0.35(\text { from }(\text { iii }) \\
\therefore \quad \mathrm{P}(\overline{\mathrm{A}} / \mathrm{B}) & =\frac{\mathrm{P}(\overline{\mathrm{A}} \cap \mathrm{B})}{\mathrm{P}(\mathrm{B})}=\frac{0.35}{0.6}=\frac{35}{60}=\frac{7}{12}
\end{aligned}
$

(v)
$
\begin{aligned}
P(\overline{\mathrm{A}} / \overline{\mathrm{B}}) & =\frac{P(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})}{\mathrm{P}(\overline{\mathrm{B}})} \\
P(\overline{\mathrm{A}} \cap \overline{\mathrm{B}}) & =P(\overline{\mathrm{A} \cup \mathrm{B}}) \\
& =1-\mathrm{P}(\mathrm{A} \cup \mathrm{B}) \\
& =1-0.65=0.35 \\
P(\overline{\mathrm{A}} / \overline{\mathrm{B}}) & =\frac{P(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})}{\mathrm{P}(\overline{\mathrm{B}})}=\frac{0.35}{0.4}=\frac{35}{40}=\frac{7}{8} .
\end{aligned}
$

Question 4.
Two cards are drawn one by one at random from a deck of 52 playing cards. What is the - probability Of getting two jacks if (i) the first card is replaced before the second card is drawn (ii) the first card is not replaced before the second card is draw?
Solution:
$
n(\mathrm{~S})=52 ; \mathrm{P}(\mathrm{J})=\frac{4}{52}
$
(i) $\mathrm{P}(\mathrm{JJ})=\frac{4}{52} \times \frac{4}{52}=\frac{1}{13} \times \frac{1}{13}=\frac{1}{169}$
(ii) $\mathrm{P}(\mathrm{JJ})=\frac{4}{52} \times \frac{3}{51}=\frac{1}{13} \times \frac{1}{17}=\frac{1}{221}$.

Question 5.
A husband and wife appear in an interview for two vacancies in the same post. The probability of husband's selection is $\frac{1}{6}$ and that of wife's selection is $\frac{1}{5}$. What is the probability that (i) both of them will be selected, (ii) only one of them will be selected, (iii) none of them will be selected?
Solution:
$
\begin{aligned}
& \mathrm{P}(\mathrm{H})=\text { Probability of husband's selection }=\frac{1}{6} \\
& \mathrm{P}(\mathrm{W})=\text { Probability of wife's selection }=\frac{1}{5}
\end{aligned}
$

and
(i)
$
\begin{aligned}
\therefore \mathrm{P}\left(\mathrm{H}^{\prime}\right) & =1-\mathrm{P}(\mathrm{H})=1-\frac{1}{6}=\frac{5}{6} \\
\mathrm{P}\left(\mathrm{W}^{\prime}\right) & =1-\mathrm{P}(\mathrm{W})=1-\frac{1}{5}=\frac{4}{5} \\
\mathrm{P}(\mathrm{HW}) & =\mathrm{P}(\mathrm{H}) \mathrm{P}(\mathrm{W})=\frac{1}{6} \times \frac{1}{5}=\frac{1}{30}
\end{aligned}
$

(ii)
$
\begin{aligned}
\mathrm{P}\left(\mathrm{HW}^{\prime} \text { or } \mathrm{WH}^{\prime}\right) & =\mathrm{P}(\mathrm{H}) \mathrm{P}\left(\mathrm{W}^{\prime}\right)+\mathrm{P}(\mathrm{W}) \mathrm{P}\left(\mathrm{H}^{\prime}\right) \\
& =\frac{1}{6} \times \frac{4}{5}+\frac{1}{5} \times \frac{5}{6}=\frac{4}{30}+\frac{5}{30} \\
& =\frac{9}{30}=\frac{3}{10}
\end{aligned}
$

(iii)
$
\mathrm{P}\left(\mathrm{H}^{\prime} \mathrm{W}^{\prime}\right)=\mathrm{P}\left(\mathrm{H}^{\prime}\right) \mathrm{P}\left(\mathrm{W}^{\prime}\right)=\frac{5}{6} \times \frac{4}{5}=\frac{4}{6}=\frac{2}{3} .
$
Question 6.
For a student the probability of getting admission in IIT is $60 \%$ and probability of getting admission in Anna university is $75 \%$. Find the probability that (i) getting admission in only one of these, (ii) getting admission in atleast one of these.

Solution:
Probability of getting admission in IIT $=\mathrm{P}(\mathrm{A})=60 \%=\frac{60}{100}=\frac{3}{5}$
$
\therefore \mathrm{P}\left(\mathrm{A}^{\prime}\right)=1-\mathrm{P}(\mathrm{A})=1-\frac{3}{5}=\frac{2}{5}
$
Probability of getting admission in Anna university $=P(B)=75 \%$
$
\begin{aligned}
& =\frac{75}{100}=\frac{3}{4} \\
\therefore P\left(B^{\prime}\right) & =1-\frac{3}{4}=\frac{1}{4}
\end{aligned}
$
(i) Probability of getting admission in only one $=\mathrm{P}\left(\mathrm{AB}^{\prime}\right.$ or $\left.\mathrm{BA}^{\prime}\right)$
$
\begin{aligned}
& =\mathrm{P}(\mathrm{A}) \mathrm{P}\left(\mathrm{B}^{\prime}\right)+\mathrm{P}(\mathrm{B}) \mathrm{P}\left(\mathrm{A}^{\prime}\right) \\
& =\frac{3}{5} \times \frac{1}{4}+\frac{3}{4} \times \frac{2}{5}=\frac{3}{20}+\frac{6}{20}=\frac{9}{20}=0.45
\end{aligned}
$
(ii) Getting admission in atleast one $=\mathrm{P}(\mathrm{A} \cup \mathrm{B})$
$
\begin{aligned}
& =P(A)+P(B)-P(A \cap B) \\
& =P(A)+P(B)-P(A) P(B) \\
& =\frac{3}{5}+\frac{3}{4}-\frac{3}{5} \times \frac{3}{4}=\frac{3}{5}+\frac{3}{4}-\frac{9}{20} \\
& =\frac{12+15-9}{20}=\frac{18}{20}=\frac{9}{10}=0.9
\end{aligned}
$