Exercise 12.4 - Chapter 12 Introduction to Probability Theory 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

$
\mathbf{E x} 12.4
$
Question 1.
A factory has two Machines-I and II. Machine-I produces $60 \%$ of items and Machine-II produces $40 \%$ of the items of the total output. Further, $2 \%$ of the items produced 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:

Question 2.
There are two identical urns containing respectively 6 black and 4 red balls, 2 black and 2 red balls. An urn is chosen at random and a ball is drawn from it, (z) find the probability that the ball is black (ii) if the ball is black, what is the probability that it is from the first urn?

Solution:

Let B be the event of getting a black ball.
Now $\quad \mathrm{P}(\mathrm{B})$ from $(\mathrm{I})=\frac{6}{10} \mathrm{P}(I)=\mathrm{P}(I I)=\frac{1}{2}$
$\mathrm{P}(\mathrm{B})$ from $(\mathrm{II})=\frac{2}{4}$
$
\begin{aligned}
\therefore \quad \mathrm{P}(\mathrm{B})= & \mathrm{P}(\mathrm{B})(\mathrm{I}) \mathrm{P}(\mathrm{I})+\mathrm{P}(\mathrm{B})(\mathrm{II}) \mathrm{P}(\mathrm{II}) \\
& =\frac{6}{10} \times \frac{1}{2}+\frac{2}{4} \times \frac{1}{2}=\frac{3}{10}+\frac{1}{4} \\
& =\frac{6+5}{20}=\frac{11}{20}
\end{aligned}
$
(ii) $\mathrm{P}(\mathrm{I} / \mathrm{B})=\frac{\mathrm{P}(\mathrm{B} / \mathrm{I}) \mathrm{P}(\mathrm{I})}{\mathrm{P}(\mathrm{B})}$
$
=\frac{\frac{6}{10} \times \frac{1}{2}}{\frac{11}{20}}=\frac{6}{20} / \frac{11}{20}=\frac{6}{11}
$
Question 3.
A firm manufactures PVC pipes in three plants viz, $\mathrm{X}, \mathrm{Y}$ and $\mathrm{Z}$. The daily production volumes from the three firms X, Y and $\mathrm{Z}$ are respectively 2000 units, 3000 units and 5000 units. It is known from the past experience that $3 \%$ of the output from plant X, $4 \%$ from plant $\mathrm{Y}$ and $2 \%$ from plant $\mathrm{Z}$ are defective. A pipe is selected at random from a day's total production,
(i) find the probability that the selected pipe is a defective one.
(ii) if the selected pipe is a defective, then what is the probability that it was produced by plant Y?
Solution:

$
\begin{aligned}
& \text { Here } \quad P(X)=\frac{2000}{10000}=\frac{2}{10} P(D / X)=\frac{3}{100} \\
& \mathrm{P}(\mathrm{Y})=\frac{3000}{1000}=\frac{3}{10} \mathrm{P}(\mathrm{D} / \mathrm{Y})=\frac{4}{100} \\
& \text { and } \mathrm{P}(\mathrm{Z})=\frac{5000}{1000}=\frac{5}{10} \mathrm{P}(\mathrm{D} / \mathrm{Z})=\frac{2}{100} \\
&
\end{aligned}
$
(i) $\mathrm{P}$ (getting a defective $\mathrm{P}($ pipe $)=\mathrm{P}(\mathrm{D})$
$
\begin{aligned}
& =P(D / X) P(X)+P(D / Y) P(Y)+P(D / Z) P(Z) \\
& =\frac{3}{100} \times \frac{2}{10}+\frac{4}{100} \times \frac{3}{10}+\frac{2}{100} \times \frac{5}{10} \\
& =\frac{6}{1000}+\frac{12}{1000}+\frac{10}{1000}=\frac{28}{1000}=\frac{7}{250}
\end{aligned}
$
(ii)
$
\mathrm{P}(\mathrm{Y} / \mathrm{D})=\frac{\mathrm{P}(\mathrm{D} / \mathrm{Y}) \mathrm{P}(\mathrm{Y})}{\mathrm{P}(D)}
$

$
=\frac{4}{100} \times \frac{3}{10} / \frac{7}{250}
$
$
=\frac{4}{100} \times \frac{3}{10} \times \frac{250}{7}=\frac{3}{7}
$
Question 4.
The changes of $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$ becoming manager of a certain company are $5: 3: 2$. The probabilities that the office canteen will be improved if $\mathrm{A}, \mathrm{B}$, and $\mathrm{C}$ become managers are $0.4,0.5$ and 0.3 respectively. If the office canteen has been improved, what is the probability that $\mathrm{B}$ was appointed as the manager?
Solution:

Given A : B : C= $5: 3: 2$
$
\begin{aligned}
\mathrm{P}(\mathrm{A}) & =\frac{5}{10} \\
\mathrm{P}(\mathrm{B}) & =\frac{3}{10} \\
\mathrm{P}(\mathrm{C}) & =\frac{2}{10} \\
\mathrm{P}(\text { Improvement of canteen } / \mathrm{B}) & =\mathrm{P}(\mathrm{I} / \mathrm{A})=\frac{4}{10} \\
\mathrm{P}(\mathrm{I} / \mathrm{B}) & =\frac{5}{10} \text { and } \mathrm{P}(\mathrm{I} / \mathrm{C})=\frac{3}{10}
\end{aligned}
$
(i) $\mathrm{P}($ Improvement of canteen $) \mathrm{P}(\mathrm{I})=\mathrm{P}(\mathrm{I} / \mathrm{A}) \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{I} / \mathrm{B}) \mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{I} / \mathrm{C}) \mathrm{P}(\mathrm{C})$
$
\begin{aligned}
& =\frac{4}{10} \times \frac{5}{10}+\frac{5}{10} \times \frac{3}{10}+\frac{3}{10} \times \frac{2}{10} \\
& =\frac{20}{100}+\frac{15}{100}+\frac{6}{100}=\frac{41}{100}
\end{aligned}
$
(ii)
$
\begin{aligned}
\mathrm{P}(\mathrm{B} / \mathrm{I}) & =\frac{\mathrm{P}(\mathrm{I} / \mathrm{B}) \mathrm{P}(\mathrm{B})}{\mathrm{P}(\mathrm{I})} \\
& =\frac{\frac{5}{10} \times \frac{3}{10}}{\frac{41}{100}}=\frac{15}{100} / \frac{41}{100}
\end{aligned}
$

Question 5.
An advertising executive is studying television viewing habits of married men and women during prime time hours. Based on the past viewing records he has determined that during prime time wives are watching television $60 \%$ of the time. It has also been determined that when the wife is watching television, $40 \%$ of the time the husband is also watching. When the wife is not watching the television, $30 \%$ of the time husband is watching the television. Find the probability that

(i) the husband is watching the television during the prime time of television

(ii) if the husband is watching the television, the wife is also watching the television.

Solution:

$\begin{aligned}
\mathrm{P}(\text { Wife watching } \mathrm{TV}) & =\mathrm{P}(\mathrm{W})=\frac{60}{100} \\
\mathrm{P}(\mathrm{H} / \mathrm{W}) & =\frac{40}{100} ; \mathrm{P}\left(\mathrm{H} / \mathrm{W}^{\prime}\right)=30 / 100
\end{aligned}$

(i)

(ii)