Exercise 12.1-Additional Questions - Chapter 12 Introduction to Probability Theory 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Problems
Question 1.
An experiment has the four possible mutually exclusive outcomes A, B, C and D. Check whether the following assignments of probability are permissible.
$
\mathrm{P}(\mathrm{A})=0.32, \mathrm{P}(\mathrm{B})=0.28, \mathrm{P}(\mathrm{C})=-0.06, \mathrm{P}(\mathrm{D})=0.46
$
Solution:
Probability of an event cannot be negative. Here $P(C)=-0.06$.
$\therefore$ the above set of events are not possible.
$
\mathrm{P}(\mathrm{A})=\frac{1}{3}, \mathrm{P}(\mathrm{B})=\frac{1}{6}, \mathrm{P}(\mathrm{C})=\frac{2}{9}, \mathrm{P}(\mathrm{D})=\frac{5}{18}
$
Question 2.
In a single throw of two dice, find the probability of obtaining
(i) sum of less than 5
(ii) a sum of greater than 10
(iii) a sum of 9 or 11.
Solution:
The sample space when throwing two dice once $=$ $\{(1,1),(1,2), \ldots \ldots \ldots \ldots(1,6)$
$(2,1)$
$(2,6)$
$
\begin{aligned}
& (6,1), \ldots \ldots \ldots(6,6)\} \\
& \mathrm{n}(\mathrm{S})=6^2=36
\end{aligned}
$
(i) Let $\mathrm{A}$ be the event of getting a sum less than 5 .
Then $A=\{(1,1),(1,2),(1,3)(2,1),(2,2)(3,1)\}$
$
\begin{aligned}
& \mathrm{n}(\mathrm{A})=6 \\
& \therefore \mathrm{P}(\mathrm{A})=\frac{n(\mathrm{~A})}{n(\mathrm{~S})}=\frac{6}{36}=\frac{1}{6}
\end{aligned}
$
(ii) Let $\mathrm{B}$ be the event of getting a sum greater than 10 .
$\therefore$ The sum will be 11 or 12 .
Now the numbers whose sum is 11 .
$
=\{(5,6),(6,5)\}
$
The number whose sum is $12=\{(6,6)\}$
$
\begin{aligned}
& \mathrm{n}(\mathrm{B})=2+1=3 \\
& \therefore \mathrm{P}(\mathrm{B})=\frac{3}{36}=\frac{1}{12}
\end{aligned}
$
(iii) Let $\mathrm{C}$ be the event of getting a sum 9 or 11 .
Now $\mathrm{C}=\{(3,6),(4,5)$
$(5,4),(6,3)$
$(5,6),(6,5)\}$
$\mathrm{n}(\mathrm{C})=6$
$
\therefore \mathrm{P}(\mathrm{C})=\frac{6}{36}=\frac{1}{6}
$

Question 3.
Three coins are tossed once. Find the probability of getting
(i) exactly two heads
(ii) at least two heads
(iii) atmost two heads.
Solution:
The sample space when three coin are tossed once is as follows:
$\mathrm{S}=\{(\mathrm{H}, \mathrm{H}, \mathrm{H}),(\mathrm{H}, \mathrm{T}, \mathrm{H}),(\mathrm{T}, \mathrm{H}, \mathrm{H}),(\mathrm{H}, \mathrm{H}, \mathrm{T}),(\mathrm{T}, \mathrm{T}, \mathrm{H}),(\mathrm{H}, \mathrm{T}, \mathrm{T})(\mathrm{T}, \mathrm{H}, \mathrm{T}),(\mathrm{T}, \mathrm{T}, \mathrm{T})\}$ $\mathrm{n}(\mathrm{S})=2^3=8$
(i) Let $\mathrm{A}$ be the event of getting exactly two heads.
$\therefore \mathrm{A}=\{(\mathrm{H}, \mathrm{T}, \mathrm{H})(\mathrm{T}, \mathrm{H}, \mathrm{H})(\mathrm{H}, \mathrm{H}, \mathrm{T})\}$
$\mathrm{n}(\mathrm{A})=3$
$\therefore \mathrm{n}(\mathrm{A})=\frac{3}{8}$
(ii) Let $B$ be the event of getting at least two heads.
$\mathrm{B}=\{(\mathrm{H}, \mathrm{T}, \mathrm{H}),(\mathrm{T}, \mathrm{H}, \mathrm{H}),(\mathrm{H}, \mathrm{H}, \mathrm{T}),(\mathrm{H}, \mathrm{H}, \mathrm{H})\}$
$\mathrm{n}(\mathrm{B})=4$
$\therefore \mathrm{P}(\mathrm{B})=\frac{4}{8}=\frac{1}{2}$
(iii) Let $C$ be the event of getting atmost two heads.
$\mathrm{C}=\{(\mathrm{T}, \mathrm{T}, \mathrm{T}),(\mathrm{H}, \mathrm{T}, \mathrm{T}),(\mathrm{T}, \mathrm{H}, \mathrm{T}),(\mathrm{T}, \mathrm{T}, \mathrm{H})(\mathrm{H}, \mathrm{H}, \mathrm{T}),(\mathrm{T}, \mathrm{H}, \mathrm{H}),(\mathrm{H}, \mathrm{T}, \mathrm{H})\}$
$\mathrm{n}(\mathrm{C})=7 \therefore \mathrm{P}(\mathrm{C})=\frac{7}{8}$
Question 4.
A bag contains 5 white and 7 black balls. 3 balls are drawn at random. Find the probability that
(i) all are white
(ii) one white and 2 black.
Solution:
Number of white balls $=5$
Number of black balls $=7$
Total number of balls $=12$
Selecting 3 from 12 balls can be done in

$
{ }^{12} \mathrm{C}_3=\frac{12 \times 11 \times 10}{3 \times 2 \times 1}=220 \text { ways }
$
$
\therefore n(\mathrm{~S})=220
$
(i) Let A be the event of selecting 3 white balls.
$
\begin{aligned}
& \therefore n(\mathrm{~A})={ }^5 \mathrm{C}_3={ }^5 \mathrm{C}_2=\frac{5 \times 4}{2 \times 1}=10 \\
& \therefore \mathrm{P}(\mathrm{A})=\frac{n(\mathrm{~A})}{n(\mathrm{~S})}=\frac{10}{220}=\frac{1}{22}
\end{aligned}
$
(ii) Let $\mathrm{B}$ be the event of selecting one white and 2 black balls.
$
\begin{aligned}
& \therefore n(\mathrm{~B})={ }^5 \mathrm{C}_1 \times{ }^5 \mathrm{C}_2=(5)\left(\frac{7 \times 6}{2 \times 1}\right)=5(21)=105 \\
& \therefore \mathrm{P}(\mathrm{B})=\frac{n(\mathrm{~B})}{n(\mathrm{~S})}=\frac{105}{220}=\frac{21}{44}
\end{aligned}
$

Question 5.
In a box containing 10 bulbs, 2 are defective. What is the probability that among 5 bulbs chosen at random, none is defective?
Solution:
Total number of bulbs $=10$
Number of defective bulbs $=2$
$\therefore$ Number of good bulbs $=10-2=8$
Now selecting 5 from the 10 bulbs can be done in ${ }^{10} \mathrm{C}_5$ ways.
(i.e.,) $n(\mathrm{~S})={ }^{10} \mathrm{C}_5=\frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1}=252$
Let $\mathrm{A}$ be the event of selecting 5 good bulbs (from 8 good bulbs)
$
\begin{aligned}
& \therefore \quad n(\mathrm{~A})=8 \mathrm{C}_5=8 \mathrm{C}_3=\frac{8 \times 7 \times 6}{3 \times 2 \times 1}=56 \\
& \therefore \quad \mathrm{P}(\mathrm{A})=\frac{n(\mathrm{~A})}{n(\mathrm{~S})}=\frac{56}{252}=\frac{2}{9}
\end{aligned}
$
Question 6.
Out of 10 outstanding students in a school there are 6 girls and 4 boys. A team of 4 students is selected at random for a quiz programme. Find the probability that there are atleast two girls.
Solution:
Let A, B and C be the three possible events of selections. The number of combinations are shown below:

$
\begin{gathered}
n(\mathrm{~S})=\text { Selecting } 4 \text { from } 10 \text { students }={ }^{10} \mathrm{C}_4=\frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1}=210 \\
n(\mathrm{~A} \cup \mathrm{B} \cup \mathrm{C})=n(\mathrm{~A})+n(\mathrm{~B})+n(\mathrm{C})
\end{gathered}
$
$
\begin{gathered}
={ }^4 \mathrm{C}_2 \times{ }^6 \mathrm{C}_2+{ }^4 \mathrm{C}_1 \times{ }^6 \mathrm{C}_3+{ }^4 \mathrm{C}_0 \times{ }^6 \mathrm{C}_4 \\
{ }^4 \mathrm{C}_2=\frac{4 \times 3}{2 \times 1}=6 ;{ }^4 \mathrm{C}_1=4 ;{ }^4 \mathrm{C}_0=1 \\
{ }^6 \mathrm{C}_2=\frac{6 \times 5}{2 \times 1}=15 ;{ }^6 \mathrm{C}_3=\frac{6 \times 5 \times 4}{3 \times 2 \times 1}=20,{ }^6 \mathrm{C}_4={ }^6 \mathrm{C}_2=15 \\
n(\mathrm{~A} \cup \mathrm{B} \cup \mathrm{C})=(6)(15)+(4)(20)+(1)(15)=90+80+15=185 \\
\therefore \quad \mathrm{P}(\mathrm{A} \cup \mathrm{B} \cup \mathrm{C})=\frac{185}{210}=\frac{37}{42} .
\end{gathered}
$
Question 7.
An integers is chosen at random from the first fifty positive integers. What is probability that the integer chosen is a prime or multiple of 4 .
Solution:
$
\mathrm{S}=\{1,2,3, \ldots \ldots \ldots, 50\} \therefore \mathrm{n}(\mathrm{S})=50
$
Let $\mathrm{A}$ be the event of getting prime number.
$
\begin{aligned}
& \therefore \mathrm{A}=\{2,3,5,7,11,13,17,19,23,29,31,37,41,43,47\} \\
& \mathrm{n}(\mathrm{A})=15 \text {, so } \mathrm{P}(\mathrm{A})=15 / 50
\end{aligned}
$
Let $B$ be the event of getting number multiple of 4
$
\begin{aligned}
& \therefore \mathrm{B}=\{4,8,12,16,20,24,28,32,36,40,44,48\} \\
& \mathrm{n}(\mathrm{B})=12, \text { so } \mathrm{P}(\mathrm{B})=12 / 50
\end{aligned}
$
Here $A$ and $B$ are mutually exclusive. (i.e.,) $A \cap B=\phi$
$
\therefore \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})=15 / 50+12 / 50=27 / 50 \text {. }
$