Exercise 12.4-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.
A factory has two Machines-I and II. Machines-I produces $25 \%$ of items and Machine-II produces $75 \%$ of the items of the total output. Further $3 \%$ of the items produces by Machine-I are defective whereas $4 \%$ produced by Machine-II are defective. If an item is drawn at random what is the probability that it is defective?
Solution:
Let $\mathrm{A}_1$ be the event that the items are produced by Machine-I and $\mathrm{A}_2$ be the event that the items are produced by Machine-II.

Let $B$ the event of drawing a defective item
$
\text { Now } \begin{aligned}
\mathrm{P}\left(\mathrm{A}_1\right) & =\frac{25}{100}=\frac{1}{4} \mathrm{P}\left(\mathrm{B} / \mathrm{A}_1\right)=\frac{3}{100} \\
\mathrm{P}\left(\mathrm{A}_2\right) & =\frac{75}{100}=\frac{3}{4} \mathrm{P}\left(\mathrm{B} / \mathrm{A}_2\right)=\frac{4}{100}
\end{aligned}
$
Now we are asked to find the total probability of the event $B$.
$
\text { (i.e.,) } \begin{aligned}
\mathrm{P}(\mathrm{B}) & =\mathrm{P}\left(\mathrm{A}_1\right) \mathrm{P}\left(\mathrm{B} / \mathrm{A}_1\right)+\mathrm{P}\left(\mathrm{A}_2\right) \mathrm{P}\left(\mathrm{B} / \mathrm{A}_2\right) \\
& =\frac{1}{4} \times \frac{3}{100}+\frac{3}{4} \times \frac{4}{100} \\
& =\frac{3}{400}+\frac{12}{400}=\frac{3+12}{400}=\frac{15}{400}=\frac{3}{80} .
\end{aligned}
$
Question 2.
There are two identical boxes containing respectively 5 white and 3 red balls, 4 white and 6 rpd balls. A box is chosen at random and a ball is drawn from it (i) find the probability that the ball is white (ii) if the ball is white, what is the probability that it from the first box?
Solution:
Let $A_1$ be the event of selecting the first box and $\mathrm{A}_2$ be the event of selecting the second box. Let $B$ be the event of selecting a white ball.

Now
$
\begin{aligned}
\mathrm{P}\left(\mathrm{A}_1\right) & =\mathrm{P}\left(\mathrm{A}_2\right)=\frac{1}{2} \\
\mathrm{P}\left(\mathrm{B} / \mathrm{A}_1\right) & =\frac{5}{8} \\
\mathrm{P}\left(\mathrm{B} / \mathrm{A}_2\right) & =\frac{4}{10}=\frac{2}{5}
\end{aligned}
$
(i)
$
\begin{aligned}
P(B) & =P\left(B / A_1\right) P\left(A_1\right)+P\left(B / A_2\right) P\left(A_2\right) \\
& =\left(\frac{5}{8}\right)\left(\frac{1}{2}\right)+\left(\frac{2}{5}\right)\left(\frac{1}{2}\right) \\
& =\frac{5}{16}+\frac{2}{10}=\frac{25+16}{80}=\frac{41}{80}
\end{aligned}
$
(ii)
$
\begin{aligned}
P\left(A_1 / B\right) & =\frac{P\left(B / A_1\right) P\left(A_1\right)}{P\left(B / A_1\right) P\left(A_1\right)+P\left(B / A_2\right) P\left(A_2\right)} \\
& =\frac{P\left(B / A_1\right) P\left(A_1\right)}{P(B)}=\frac{\left(\frac{5}{8}\right)\left(\frac{1}{2}\right)}{\frac{41}{80}} \\
& =\frac{\frac{5}{46}}{\frac{41}{80}}=\frac{5}{16} \times \frac{80}{41}=\frac{5 \times 5}{41}=\frac{25}{41}
\end{aligned}
$

Question 3.
In a factory, Machine-I produces $45 \%$ of the output and Machine-II produces $55 \%$ of the output. On the average $10 \%$ items produced by I and $5 \%$ of the items produced by II are defective. An item is drawn at random from a day's output, (i) Find the probability that it is a defective item (ii) If it is defective, what is the probability that it was produced by Machine-II?
Solution:
Let $\mathrm{A}_1$ and $\mathrm{A}_2$ be the events that the items produced by Machine-I and II respectively.

Let $B$ be the event of selecting a defective item
$
\begin{aligned}
\mathrm{P}\left(\mathrm{A}_1\right) & =\frac{45}{100}=\frac{9}{20} ; \quad \mathrm{P}\left(\mathrm{A}_2\right)=\frac{55}{100}=\frac{11}{20} \\
\mathrm{P}\left(\mathrm{B} / \mathrm{A}_1\right) & =\frac{10}{100}=\frac{1}{10} \\
\mathrm{P}\left(\mathrm{B} / \mathrm{A}_2\right) & =\frac{5}{100}=\frac{1}{20}
\end{aligned}
$
(i) $\mathrm{P}(\mathrm{B})=\mathrm{P}\left(\mathrm{B} / \mathrm{A}_1\right) \mathrm{P}\left(\mathrm{A}_1\right)+\mathrm{P}\left(\mathrm{B} / \mathrm{A}_2\right) \mathrm{P}\left(\mathrm{A}_2\right)$
$
=\frac{1}{10} \times \frac{9}{20}+\frac{1}{20} \times \frac{11}{20}=\frac{9}{200}+\frac{11}{400}=\frac{18+11}{400}=\frac{29}{400}
$

(ii)
$
\begin{aligned}
\mathrm{P}\left(\mathrm{A}_2 / \mathrm{B}\right) & =\frac{\mathrm{P}\left(\mathrm{B} / \mathrm{A}_2\right) \mathrm{P}\left(\mathrm{A}_2\right)}{\mathrm{P}\left(\mathrm{B} / \mathrm{A}_1\right) \mathrm{P}\left(\mathrm{A}_1\right)+\mathrm{P}\left(\mathrm{B} / \mathrm{A}_2\right) \mathrm{P}\left(\mathrm{A}_2\right)}=\frac{\mathrm{P}\left(\mathrm{B} / \mathrm{A}_2\right) \mathrm{P}\left(\mathrm{A}_2\right)}{\mathrm{P}(\mathrm{B})} \\
& =\frac{\frac{1}{20} \times \frac{11}{20}}{\frac{29}{400}}(\text { from }(i)) \\
& =\frac{11 / 400}{29 / 400}=\frac{11}{29}
\end{aligned}
$