Exercise 9.2 - Chapter 9 Probability 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Eх $9.2$
Question $1 .$
A company manufactures 10000 Laptops in 6 months. Out of which 25 of them are found to be defective. When you choose one Laptop from the manufactured, what is the probability that selected Laptop is a good one.
Solution:
Total $n(S)=10,000$
Defective $\mathrm{n}(\mathrm{A})=25$

No. of good laptops $=1000-25$
$\mathrm{n}(\mathrm{B})=9975$

Question $2 .$
In a survey of 400 youngsters aged 16-20 years, it was found that 191 have their voter ID card. If a youngster is selected at random, find the probability that the youngster does not have their voter ID card.
Solution:
No. of youngsters $n(S)=400$
No. of youngsters having voter id $n(A)=191$
No. of youngsters do not have their voter id $n(B)=400-191=209$
$\mathrm{P}(\mathrm{B})=\frac{n(\mathrm{~B})}{n(\mathrm{~S})}=\frac{209}{400}$

Question $3 .$
The probability of guessing the correct answer to a certain question is $\frac{x}{3}$. If the probability of not guessing the correct answer is $\frac{x}{5}$, then find the value of $x$.
Solution:
$\begin{aligned}
\frac{x}{3}+\frac{x}{5} &=1 \\
\frac{5 x+3 x}{15} &=1 \\
\frac{8 x}{15} &=1 \\
8 x &=15 \\
x &=\frac{15}{8}
\end{aligned}$

Question $4 .$
If a probability of a player winning a particular tennis match is $0.72$. What is the probability of the player loosing the match?
Solution:
$P(A)=0.72$
$P\left(A^{\prime}\right)=1-0.72=0.28$

Question $5 .$
1500 families were surveyed and following data was recorded about their maids at homes

A family is selected at random. Find the probability that the family selected has
(i) Both types of maids
(ii) Part time maids
(iii) No maids
Solution:
$\mathrm{n}(\mathrm{S})=1500$ (Total families)
$\mathrm{n}(\mathrm{A})=860$ (Part time maids)
$n(B)=370$ (Only full time)
$n(A \cap B)=250($ Both $)$
(i)
$\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{n(\mathrm{~A} \cap \mathrm{B})}{n(\mathrm{~S})}=\frac{250}{1500}=\frac{1}{6}$
(ii)
$\mathrm{P}(\mathrm{A})=\frac{n(\mathrm{~A})}{n(\mathrm{~S})}=\frac{.860}{1500}=\frac{43}{75}$

Total families $n(S)=1500$
No. of families have maids $=860+250+370=1480$
No. of families do not have maids $=1500-1480$ $\mathrm{n}(\mathrm{A})=20$
$\therefore \mathrm{P}(\mathrm{A})=\frac{n(\mathrm{~A})}{n(\mathrm{~S})}=\frac{20}{1500}=\frac{1}{75}$