Search
You can find the results of your search below.
Fulltext results:
- Question 13 Exercise 6.2
- word "Excellence." How many of these permutations begin with $\mathrm{E}$ ? ====Solution==== The total nu... m_2=2$ are $L$ and $m_3=2$ are $C$. Therefore, \begin{align}\text{total number of permutations are} &=\left(\begin{array}{c} n \\ m_1, m_2, m_3 \end{array}\right)\\&=\left(\begin{array}{c} 10 \\ 4,2,2 \end{array}\right) \\ & =\d
- Question 7 Exercise 6.4
- n==== The sample space rolling a pair of dice is \begin{align}S&=\{(i, j) ; i, j=1,2,3,4,5,6\}\\ &=\left[\begin{array}{llllll} (1,1) & (1,2) & (1,3) & (1,4) & (1... )=6 \times 6=36$$ doublet of even numbers. Let \begin{align}A&=\{(2,2),(4,4),(6,6)\}\\ n(A)&=3\end{alig... n==== The sample space rolling a pair of dice is \begin{align} S&=\{(i, j) ; i, j=1,2,3,4,5,6\}\\ &=\left
- Question 9 Exercise 6.3
- ill exactly contain four men and four women are: \begin{align}{ }^7 C_4 \cdot{ }^6 C_4&=\dfrac{7 !}{(7-4)... in this case the total number of committees are: \begin{align}{ }^7 C_2 \cdot{ }^6 C_6&=\dfrac{7 !}{(7-2)... in this case the total number of committees are: \begin{align}{ }^7 C_3 \cdot{ }^6 C_5&=\dfrac{7 !}{(7-3)... n, in this case total number of committees are: \begin{align}{ }^7 C_4 \cdot{ }^6 C_4&=\dfrac{7 !}{(7-4)
- Question 12 Exercise 6.2
- mber of different words using all at a time are: \begin{align} \left(\begin{array}{c} n \\ m 1 \end{array}\right)&=\left(\begin{array}{l} 8 \\ 3 \end{array}\right) \\ & =\dfrac{8 !... $. Thus the total number of different words are: \begin{align} \left(\begin{array}{c} n \\ m_1, m_2, m_3
- Question 4 Exercise 6.4
- Hence the sample space of the given problem is: \begin{align}S&=(HHII,HHT.HTH.HTT.THII.THT.TTH,TT7),\\ \... Hence the sample space of the given problem is: \begin{align}S&=(HHII,HHT.HTH.HTT.THII.THT.TTH,TT7),\\ ... Hence the sample space of the given problem is: \begin{align}S&=(HHII,HHT.HTH.HTT.THII.THT.TTH,TT7),\\ ... Hence the sample space of the given problem is: \begin{align}S&=(HHII,HHT.HTH.HTT.THII.THT.TTH,TT7),\\ n
- Question 1 and 2 Exercise 6.1
- frac{10 !}{3 ! .3 ! \cdot 4 !}$ ====Solution==== \begin{align}\dfrac{10 !}{3 ! \cdot 3 ! \cdot 4 !}&=\dfr... the $\dfrac{3 !+4 !}{5 !-4 !}$ ====Solution==== \begin{align}\dfrac{3 !+4 !}{5 !-4 !}&=\dfrac{3 !+4.3 !}... the $\dfrac{(n-1) !}{(n+1) !}$ ====Solution==== \begin{align}\dfrac{(n-1) !}{(n+1) !} &= \dfrac{(n-1) !... ate the $\dfrac{10 !}{(5 !)^2}$ ====Solution==== \begin{align}\dfrac{10 !}{(5 !)^2}&=\dfrac{10 \cdot 9 \c
- Question 1 and 2 Exercise 6.5
- ution==== We know by addition law of probability \begin{align} P(A \cup B)&=P(A)+P(B)-P(A \cap B) \\ \R... mplementary events $$P(B)=1-P(\bar{B})$$ Putting \begin{align}P(\bar{B})&=\dfrac{5}{8}\\ P(B)&=1-\dfrac{5... })=1-P(A)$$ Putting $P(A)=\dfrac{1}{2}$, we get \begin{align}P(\bar{A})&=1-\dfrac{1}{2}=\dfrac{1}{2}\\ P... w by addition law of probability, we know that: \begin{align}P(A \cap B)&=P(A)+P(B)-P(A \cap B) \\ \Righ
- Question 9 Exercise 6.5
- ity that both will be selected. ====Solution==== \begin{align} P(\text { Ajmal scicction })&=\dfrac{1}{7}... therefore these two events are independent. Thus \begin{align}P(\text { Both are selected })&=P(A) \times... lity that Only one is selected. ====Solution==== \begin{align} P(\text { Ajmal scicction })&=\dfrac{1}{7}... t events. Thus the probability that only one is \begin{align} \text { selected is: }&=P(A \cap \bar{B})+
- Question 1 and 2 Exercise 6.2
- ion 1(i)===== Evaluate $^6 P_6$ ====Solution==== \begin{align}^6 P_6&=\dfrac{6 !}{(6-6) !}\\ &=6 !=720\en... 1(ii)===== Evaluate $^{20} P_2$ ====Solution==== \begin{align}^{20} P_2&=\dfrac{20 !}{(20-2) !}\\ &=\dfra... (iii)===== Evaluate $^{16} P_3$ ====Solution==== \begin{align}^{16} P_3&=\dfrac{16 !}{(16-3) ! }\\ &=\dfr... ^n P_3)$ for $n.$ ====Solution==== We are given: \begin{align}^n P_5&=56(^n P_3) \\ \Rightarrow \dfrac{n
- Question 2 Exercise 6.3
- d ${ }^n C_r=35$. ====Solution==== We are given: \begin{align} &^n P_r=\dfrac{n !}{(n-r) !}=840 ....(i)\\... 5....(ii)\end{align} Dividing Eq.(i) by Eq.(ii) \begin{align}\dfrac{n !}{(n-r) !} \cdot \dfrac{(n-r) ! r... &=4\end{align} Putting $r=4$ in Eq.(ii), we get \begin{align} & { }^n C_4=\dfrac{n !}{(n-4) ! 4 !}=35 \\... $y=n^2-3 n$ then the above last equation becomes \begin{align} & y(y+2)=840 \\ & \Rightarrow y^2+2 y-840=
- Question 5 Exercise 6.1
- on==== We are taking L.H.S of the above equation \begin{align}\dfrac{(2 n) !}{n !}&=\dfrac{1}{n !}[(2 n)(... .H.S of the above equation are total $2 n$ terms \begin{align}\dfrac{(2 n) !}{n !}&=\dfrac{1}{n !}[(2 n)(... kets contain $n$ terms, that can be simplify as: \begin{align}\dfrac{(2 n) !}{n !}& =\dfrac{1}{n !}[2.2 .... on==== We are taking L.H.S of the above equation \begin{align}\dfrac{(2 n+1) !}{n !}&=\dfrac{1}{n !}[(2 n
- Question 1 Exercise 6.4
- lling a $5$ ? ====Solution==== Rolling a $5$ Let \begin{align}A&=\{5\}\\ P(A)&=\dfrac{n(A)}{n(S)}\\ &=\df... =Solution==== Rolling a number less than $1$ Let \begin{align}B&=\{\}\\ &=\phi \text{then}\\ P(B)&=\dfrac... lution==== Rolling a number greater than $0$ Let \begin{align}C&=\{1,2,3,4,5,6\},\text{then}\\ P(C)&=\dfr... ? ====Solution==== Rolling a multiple of $3$ Let \begin{align}D&=\{3,6\}\text{then}\\ P(D)&=\dfrac{n(D)}{
- Question 3 and 4 Exercise 6.5
- exclusive, therefore $A \cap B=\emptyset$. Thus \begin{align}P(A \cup B)&=P(A)+P(B)\\ \Rightarrow P(B)&=... olution==== Total numbers written on tickets are \begin{align}S&=\{1,2,3, \ldots, 50\} \text { so }\\ n(S)&=50 \end{align} Let \begin{align}A \{odd \,numbers \}&=\{1,3,5,..,29\}\\ n(A... \}\\ &=\{1,4,9,16,25\}\\ n(B)&=5\end{align} Álso \begin{align}A \cap B&=\{1,9,25\}\\ \text{Therefore} n(A
- Question 3 & 4 Exercise 6.1
- re taking the L.H.S of the above given equation. \begin{align}\dfrac{1}{6 !}+\dfrac{2}{7 !}+\dfrac{3}{8 !... re taking the L.H.S of the above given equation. \begin{align}\dfrac{(n+5) !}{(n+3) !}&=\dfrac{(n+5)(n+4)... 2(n !)}{(n-4) !}$ ====Solution==== We are given: \begin{align}\dfrac{n(n !)}{(n-5) !}&=\dfrac{12(n !)}{(n... !}{(n-4) !}=9: 1$ ====Solution==== We are given: \begin{align} \dfrac{n !}{(n-4) !}: \dfrac{(n-1) !}{(n-4
- Question 10 Exercise 6.2
- r of ways these five students can be seated are: \begin{align}^8 P_5&=\dfrac{8 !}{(8-5) !}\\ &=\dfrac{8 \... s 7. In this case the total number of ways are: \begin{align}^2 P_2 \times^7 P_4&=2 \times \dfrac{7 !}{(... r of ways these five students can be seated are: \begin{align}^8 P_5&=\dfrac{8 !}{(8-5) !}\\ &=\dfrac{8 \... number ways sitting these students in a row are: \begin{align}^7 P_4&= \dfrac{7 !}{(7-4) !}\\ &=\dfrac{7.