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- Question 1 and 2 Exercise 6.5
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... $.\\ Find $P(A \cap B)$. ====Solution==== We know by addition law of probability \begin{align} P(A \c... =\dfrac{5}{8}, P(A \cup B)=\dfrac{3}{4}$$ We know by complementary events $$P(B)=1-P(\bar{B})$$ Puttin... r{B})&=1-\dfrac{1}{2}=\dfrac{1}{2}\end{align} Now by addition law of probability, we know that: \begi
- Question 7 and 8 Exercise 6.2
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... git, ten digit and unit digit place) to be filled by five digits, Moreover repetition is allowed. H... t m_2=5$ ways $E_3$ occurs in $m_3=5$ ways Thus by fundamental principle of counting the total numbe... $m_2=4$ ways $E_3$ occurs in $m_3=3$ ways. Thus by fundamental principle of 'counting the total numb
- Question 5 and 6 Exercise 6.2
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... umber of possible arrangements to fill $7$ places by these $7$ letters are: \begin{align}^7 P_7&=\dfra... r even number, the unit digit have to be filled by $2$ or $4$. So, we are left with $3$ numbers. Th... =2$ Ten Thousand: $E_4$ occurs in $m_4=1$. Thus by fundamental principle of counting the total numbe
- Question 11 Exercise 6.2
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... $E_2$ occurs in $m_2=5$. Hence the total numbers by fundamental principle of counting greater than $1... digits but hundred digit place can not be occupy by the digit $0$ \\ Hundred digit: Event $E_1$ occur... Unit digit:Event $E_3$ occurs in $m_3=4$. Hence by fundamental principle of counting numbers greater
- Question 3 and 4 Exercise 6.2
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... eshawar, Pakistan. =====Question 3(i)===== Prove by Fundamental principle of counting $^n P_r=n(^{n-1... ^n P_r\end{align} =====Question 3(ii)===== Prove by Fundamental principle of counting $^n P_r=^{n-1}
- Question 5 and 6 Exercise 6.3
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... 5(i)===== How many straight lines are determined by $12$ points, no three of which lie on the same st... tion 5(ii)===== How many triangles are determined by $12$ points, no three of which lie on the same st
- Question 5 Exercise 6.4
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... n(S)&=252\end{align} When $3$ men and $2$ women. By multiplication principle the total number of ways... (S)&=252\end{align} When $2$ men and $3$ wornen. By multiplication principle the total number of ways
- Question 7 Exercise 6.5
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... ince the events are mutually disjoints therefore, by addition law of probability we have \begin{align}... n} Now the card should be neither red nor king so by complementary events, we have that probability t
- Question 8 Exercise 6.5
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... is $7$ or $11$ are mutually exclusive, therefore by addition law of probability we have \begin{alig... &=\dfrac{8}{36}\\ &=\dfrac{2}{9}\end{align} Thus by complementary events, the probability that sum of
- Question 2 Review Exercise 6
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... n-(r+2)) !(r+2) !}\end{align} Dividing both sides by $(2 n)$ ! we get \begin{align} \Rightarrow \dfrac... 8-(r+2)] !(r+2) !}\end{align} Dividing both sides by $18!$ \begin{align}\dfrac{1}{(18-r) ! r !}&= \dfr
- Question 5 & 6 Review Exercise 6
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... ot 2 !=48$$ Because the five people will permute by $(5-1) !$ and Faisal and Saima can permute by 2 ! ====Go To==== <text align="left"><btn type="prim
- Question 11 Review Exercise 6
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... ds for red event and $G$ stands for green event. By addition law of probability, we have \begin{align... of red is: $$P(\text { Red })=\dfrac{1}{4}$$ Then by complementary event theorem: \begin{align} P(\tex
- Question 9 Exercise 6.2
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... ====Question 9===== How many signals can be given by six flags of different colors when any number of
- Question 2 Exercise 6.3
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... n-r) ! r !}=35....(ii)\end{align} Dividing Eq.(i) by Eq.(ii) \begin{align}\dfrac{n !}{(n-r) !} \cdot
- Question 9 Exercise 6.3
- Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB)... that four women to be selected are ${ }^6 C_4$. By fundamental principle of counting the total numbe