Exercise 9.1 - Chapter 9 Probability 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Ex $9.1$ : Chapter 9 - Probability - 9th Maths Guide Samacheer Kalvi Solutions
Question 1.
You are walking along a street. If you just choose a stranger crossing you, what is the probability that his next birthday will fall on a Sunday?
Solution:
Days in a week (S) $=\{$ Sunday, Monday, Tuesday, Wednesday, Thursday,
Friday, Saturday $\}$
$\mathrm{n}(\mathrm{S})=7$
$\therefore$ No. of days in week $=7$
Event of selecting Sunday $(\mathrm{A})=\{$ Sunday $\}$
$\mathrm{n}(\mathrm{A})=1$
$\therefore$ Probability of selecting Sunday $=\frac{n(\mathrm{~A})}{n(\mathrm{~S})}=\frac{1}{7}$

Question 2.
What is the probability of drawing a King or a Queen or a Jack from a deck of cards?
Solution:
Number of cards $n(S)=52$
No. of King cards $n(A)=4$
No. of Queen cards $n(B)=4$
No. of Jack cards $n(C)=4$
Probability of drawing a King card

$=\frac{n(\mathrm{~A})}{n(\mathrm{~S})}=\frac{4}{52}$
Probability of drawing a Queen card
$=\frac{n(\mathrm{~B})}{n(\mathrm{~S})}=\frac{4}{52}$
Probability of drawing a Jack card
$=\frac{n(\mathrm{C})}{n(\mathrm{~S})}=\frac{4}{52}$
The Probability of drawing a King or a Queen or a Jack from a deck of cards $=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{C})=\frac{4}{52}+\frac{4}{52}+\frac{4}{52}=\frac{4+4+4}{52}=\frac{12}{52}=\frac{3}{13}$

Question $3 .$
What is the probability of throwing an even number with a single standard dice of six faces?
Solution:

Faces of a dice $(S)=\{1,2,3,4,5,6\}$
$n(S)=6$
Event of throwing an even number
$\mathrm{A}=\{2,4,6\}, \mathrm{n}(\mathrm{A})=3$
$\square$ Probability of throwing an even number
$\mathrm{P}(\mathrm{A})=\frac{n(\mathrm{~A})}{n(\mathrm{~S})}=\frac{3}{6}=\frac{1}{2}$
 

Question $4 .$
There are 24 balls in a pot. If 3 of them are Red, 5 of them are Blue and the remaining are Green then, what is the probability of picking out
(i) a Blue ball,
(ii) a Red ball and
(iii) a Green ball?
Solution:
$\mathrm{n}(\mathrm{S})=24$
Red $-n(R)=3$
Blue $-n(B)=5$
Green $-n(G)=16$
(i)
Probability of picking a Blue ball $=\frac{n(\mathrm{~B})}{n(\mathrm{~S})}=\frac{5}{24}$
(ii)
Probability of picking a Red ball $=\frac{n(\mathrm{R})}{n(\mathrm{~S})}=\frac{3}{2_{8}^{24}}=\frac{1}{8}$
(iii)
Probability of picking a Green ball $=\frac{n(\mathrm{G})}{n(\mathrm{~S})}=\frac{2_{2}^{2}}{24}=\frac{2}{3}$

Question $5 .$
When two coins are tossed, what is the probability that two heads are obtained?
Solution:
Sample space when two coins are tossed $(\mathrm{S})=\{\mathrm{HH}, \mathrm{TT}, \mathrm{HT}, \mathrm{TH}\}$
$\mathrm{n}(\mathrm{S})=4$
Event of getting two heads $(\mathrm{A})=\{\mathrm{HH}\}$
$\mathrm{n}(\mathrm{A})=1$
Probability of getting two heads $\mathrm{P}(\mathrm{A})=\frac{n(\mathrm{~A})}{n(\mathrm{~S})}=\frac{1}{4}$
 

Question $6 .$
Two dice are rolled, find the probability that the sum is
(i) equal to 1
(ii) equal to 4
(iii) less than 13
Solution:
When two dice are rolled Sample space

$\begin{aligned}
&\mathrm{S}=\left\{\begin{array}{l}
(1,1),(1,2),(1,3),(1,4),(1,5),(1,6), \\
(2,1),(2,2),(2,3),(2,4),(2,5),(2,6), \\
(3,1),(3,2),(3,3),(3,4),(3,5),(3,6), \\
(4,1),(4,2),(4,3),(4,4),(4,5),(4,6), \\
(5,1),(5,2),(5,3),(5,4),(5,5),(5,6), \\
(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)
\end{array}\right\} \\
&n(\mathrm{~S})=36
\end{aligned}$
(i) Event of the sum is equal to $1=0$
$\therefore$ Probability $=\frac{0}{n(\mathrm{~S})}=0$
(ii) Event of the sum is equal to 4
$\begin{aligned}
\mathrm{B} &=\{(1,3),(2,2),(3,1)\} \\
n(\mathrm{~B}) &=3 \\
\mathrm{P}(\mathrm{B}) &=\frac{n(\mathrm{~B})}{n(\mathrm{~S})}=\frac{\not \beta}{36}=\frac{1}{12}
\end{aligned}$
(iii) Event of the sum is equal to less than 13
$C=\left\{\begin{array}{l}
(1,1),(1,2),(1,3),(1,4),(1,5),(1,6), \\
(2,1),(2,2),(2,3),(2,4),(2,5),(2,6), \\
(3,1),(3,2),(3,3),(3,4),(3,5),(3,6), \\
(4,1),(4,2),(4,3),(4,4),(4,5),(4,6), \\
(5,1),(5,2),(5,3),(5,4),(5,5),(5,6), \\
(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)
\end{array}\right\}$

$n(\mathrm{C})=36$
$\mathrm{P}(\mathrm{C})=\frac{n(\mathrm{C})}{n(\mathrm{~S})}=\frac{36}{36}=1$

Question $7 .$
A manufacturer tested 7000 LED lights at random and found that 25 of them were defective. If a LED light is selected at random, what is the probability that the selected LED light is a defective one.
Solution:
$n(S)=7000 \mathrm{~S}$ - Total no. of lights.
$\mathrm{n}(\mathrm{A})=25 \mathrm{~A}-$ Defective ones.
$\mathrm{P}(\mathrm{A})=\frac{n(\mathrm{~A})}{n(\mathrm{~S})}=\frac{25}{\frac{2000}{280}}=\frac{1}{280}$

Question $8 .$
In a football match, a goalkeeper of a team can stop the goal, 32 times out of 40 attempts tried by a team. Find the probability that the opponent team can convert the attempt into a goal.

Solution

Total no. of attempts $n(S)=40$
Total no. of attempts by A team $n(A)=32$
Total no. of attempts by the opponent team $B=n(B)=40-32=8$

Question $9 .$
What is the probability that the spinner will not land on a multiple of 3 ?
Solution:
Total no. of choices $n(S)=8$
Total no. of multiples of $3 \mathrm{~A}=\{3,6\}$
$\mathrm{n}(\mathrm{A})=2$
Event of non-multiples of $3 \mathrm{~B}=\{1,2,4,5,7,8\}$
$n(B)=6$
$\therefore \mathrm{P}(\mathrm{B})=\frac{n(\mathrm{~B})}{n(\mathrm{~S})}=\frac{\frac{3}{8}}{\frac{8}{4}}=\frac{3}{4}$

Question $10 .$
Frame two problems in calculating probability, based on the spinner shown here.

Solution:
(i) What is the probability that the spinner will land on an even number?
(ii) What is the probability that the spinner will not land on a prime number.