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- Question 9 Exercise 6.3
- n==== Total men are $7$ and total women are $6.$ Therefore, Total number of persons $=7+6=13$ Committee... . In how many ways can the committee be chosen if there must be at least two men? ====Solution==== Total men are $7$ and total women are $6$. Therefore, Total number of persons $=7+6=13$ There must be at least two men The number of committees that wil
- Question 13 Exercise 6.2
- 1=4$ are $E, m_2=2$ are $L$ and $m_3=2$ are $C$. Therefore, \begin{align}\text{total number of permutat... 3$ are $E$. $m_2=2$ are $L$ and $m_3=2$ are $C$. Therefore, \begin{align}\text{Number of permulations a... $ are $E$, $m_2=2$ are $L$ $m_3=2$ are $C$. Therefore, \begin{align}\text{total number of permutat... are $E$, $m_2=2$ are $L$ and $m_3=1$ are $C$. $\therefore$ \begin{align}\text{Number of permutations a
- Question 3 Exercise 6.4
- ve $8$ questions, each question has two options. Therefore, The state space contains $2^8$ distinct outc... ay i.e. $${ }^8 C_8=\dfrac{8 !}{(8-8) ! 8 !}=1$$ Therefore probability to $8$ answers are correct is: $$... ve $8$ questions, each question has two options. Therefore, The state space contains $2^8$ distinct outc... ve $8$ questions, each question has two options. Therefore, The state space contains $2^8$ distinct outc
- Question 7 and 8 Exercise 6.2
- 4$ and 5 if repetitions allowed? ====Solution==== There are three places (hundred digit, ten digit and un... ll deal all the vowels as a single alphabet. So, there are four places to be filled with four alphabets.
- Question 1 Exercise 6.3
- or } n=-8\end{align} But $n$ can not be negative therefore, we have $n=9$. =====Question 1(ii)===== Sol... or } n=8 \end{align} But $n$ can not be negative, therefore we have $n=8$. =====Question 1(iii)===== So
- Question 5 and 6 Exercise 6.3
- any line twice. the the answer is "No," because there are no three of the given points on any line. ==... ny triangle twice. Again, the answer is "No." If there were a fourth point in one of these triangles,
- Question 3 and 4 Exercise 6.5
- that $\mathrm{A}$ and $B$ are mutually exclusive, therefore $A \cap B=\emptyset$. Thus \begin{align}P(A ... n} Álso \begin{align}A \cap B&=\{1,9,25\}\\ \text{Therefore} n(A \cap B)&=3\end{align} Now $$P(A)=\dfrac{
- Question 9 Exercise 6.5
- n of one does not depend on the other sclection, therefore these two events are independent. Thus \begin... {5}\end{align} Only one is selected In this case there are two chance. it may be Ajmal selected Bushra
- Question 11 Review Exercise 6
- ccvers one fourth $\dfrac{1}{4}$ of the spinner. Therefore, \begin{align}P(\operatorname{Red})&=\dfrac{1... n} Also these two are mutually exclusive events. Therefore $P(R \cap G)=\phi$, where $R$ stands for red
- Question 3 & 4 Exercise 6.1
- ext { or } n=6\end{align} $n$ can not b negative, therefore $n=6$. =====Question 4(ii)===== Find the val
- Question 1 and 2 Exercise 6.2
- $n$ can not be negative nor it can be imaginary, therefore, $n=5$. ====Go To==== <text align="righ
- Question 7 and 8 Exercise 6.3
- veryone will shake hand with every other person, therefore the total number of shake hands are: \begin{
- Question 5 and 6 Exercise 6.5
- ary events, and we are given $P(E)=\dfrac{8}{9}$, therefore, \begin{align}P(E^{\prime})&=1-P(E)=1-\dfrac{
- Question 7 Exercise 6.5
- {1}{13}$$ Since the events are mutually disjoints therefore, by addition law of probability we have \begi
- Question 8 Exercise 6.5
- t the sum is $7$ or $11$ are mutually exclusive, therefore by addition law of probability we have \beg