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- Question 13 Exercise 6.2
- re} &=\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) \\ & =\dfrac{10 !}{4 ! \cdot 2 ! \cdot 2 !}\\ &=37,800 \end{align} Begin with $\mathrm{E}$ If we have to pic... are} &=\left(\begin{array}{c} n \\ m_1, m_2, m_3 \end{array}\right)\\&=\dfrac{9 !}{3 ! \cdot 2 ! \cdot
- Question 7 Exercise 6.4
- \\ (6.1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6) \end{array}\right]\end{align} So total number of sample points are $$n(S)=6 \times 6=36$$ doublet of even ... t \begin{align}A&=\{(2,2),(4,4),(6,6)\}\\ n(A)&=3\end{align} Hence the possibility of getting doublet o... \\ (6.1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6) \end{array}\right]\end{align} So total number of sampl
- Question 9 Exercise 6.3
- }{(7-4) ! 4 !} \cdot \dfrac{6 !}{(6-4)}\\\ &= 525\end{align} =====Question 9(ii)===== An $8$-persons c... 7-2) ! 2 !} \cdot \dfrac{6 !}{(6-6) ! 6 !}\\ &=21\end{align} If committee contains $3$ men then it wil... 3) ! 3 !} \cdot \dfrac{6 !}{(6-5) ! 5 !}\\\ &=210\end{align} If committee contain $4$ men then it will ... 4) ! 4 !} \cdot \dfrac{6 !}{(6-4) ! 4 !}\\\ &=525\end{align} If committee contains $5$ men then it will
- Question 12 Exercise 6.2
- : \begin{align} \left(\begin{array}{c} n \\ m 1 \end{array}\right)&=\left(\begin{array}{l} 8 \\ 3 \end{array}\right) \\ & =\dfrac{8 !}{3 !}\\ &=\dfrac{8 \c... cdot 6 \cdot 5 \cdot 4 \cdot 3 !}{3 !}\\ &=6,720 \end{align} =====Question 12(ii)===== How many differ... align} \left(\begin{array}{c} n \\ m_1, m_2, m_3 \end{array}\right)&=\left(\begin{array}{c} 10 \\ 2,3.2
- Question 4 Exercise 6.4
- .HTT.THII.THT.TTH,TT7),\\ \text{then} n(S)&=2^3=8\end{align} When all heads. Let $$A=\{H H H\}$$ then ... HTT.THII.THT.TTH,TT7),\\ \text{then} n(S)&=2^3=8\end{align} When two heads Let $$B=\{H T$ T.THT.TTH $... HII,HHT.HTH.HTT.THII.THT.TTH,TT7),\\ n(S)&=2^3=8\end{align} When one head. Let $$C=\{H H T, H T H, T ... HHII,HHT.HTH.HTT.THII.THT.TTH,TT7),\\ n(S)&=2^3=8\end{align} When at least one head. Let \begin{align}
- Question 1 and 2 Exercise 6.1
- 4 !}\\ &=\dfrac{10.9 .8 .7 .5}{3.2 .1}\\ &=4200 \end{align} =====Question 1(ii)===== Evaluate the $\d... \dfrac{3 !(5)}{4 \cdot 3 !(4)}\\ &=\dfrac{5}{16} \end{align} =====Question 1(iii)===== Evaluate the $\... {(n-1) !}{(n+1) n(n-1) !} \\ &= \dfrac{1}{n(n+1)}\end{align} =====Question 1(iv)===== Evaluate the $\d... cdot 4 \cdot 3 \cdot 2 \cdot 1 \cdot 5 !}\\ &=252\end{align} =====Question 2(i)===== Write $19.18.17 .
- Question 1 and 2 Exercise 6.5
- \Rightarrow P(A \cap B)&=P(A)+P(B)-P(A \cup B) \end{align} Substituting $P(A), P(B)$ and $P(A \cup B)... =\dfrac{5}{8}\\ P(B)&=1-\dfrac{5}{8}=\dfrac{3}{8}\end{align} Also $$P(\bar{A})=1-P(A)$$ Putting $P(A)=... bar{A} \cap \bar{B})&=1-\dfrac{1}{2}=\dfrac{1}{2}\end{align} Now by addition law of probability, we kno... \\ \Rightarrow P(A \cap B)&=P(A)+P(B)-P(A \cup B)\end{align} Using the above given and calculated \begi
- Question 9 Exercise 6.5
- ow P(\text { Bushra not selected })&=\dfrac{4}{5}\end{align} Both are selected Since the selection of ... & =\dfrac{1}{7} \times \dfrac{1}{5}=\dfrac{1}{35}\end{align} =====Question 9(ii)===== A'jmal and Bushr... ow P(\text { Bushra not selected })&=\dfrac{4}{5}\end{align} Only one is selected In this case there a... t \dfrac{1}{5} \\ & =\dfrac{10}{35}=\dfrac{2}{7} \end{align} =====Question 9(iii)===== A'jmal and Bush
- Question 1 and 2 Exercise 6.2
- in{align}^6 P_6&=\dfrac{6 !}{(6-6) !}\\ &=6 !=720\end{align} =====Question 1(ii)===== Evaluate $^{20} ... &=\dfrac{20.19 .18 !}{18 !}\\ &=20 \times 19=380\end{align} =====Question 1(iii)===== Evaluate $^{16} ... 16 \cdot 15 \cdot 14 \cdot 13 !}{13 ! }\\ &=3360 \end{align} =====Question 2(i)===== Solve $^n P_5=56(... htarrow(n-11)(n+4)&=0\\ n=11& \text{or}\quad n=-4\end{align} But $n$ can not be negative, so $n=11$. =
- Question 2 Exercise 6.3
- (i)\\ &^n C_r=\dfrac{n !}{(n-r) ! r !}=35....(ii)\end{align} Dividing Eq.(i) by Eq.(ii) \begin{align}\... &=\dfrac{840}{35}\\ r!&=24\\ \text{or}\quad r &=4\end{align} Putting $r=4$ in Eq.(ii), we get \begin{al... -2)=840 \\ & \Rightarrow(n^2-3 n)(n^2-3 n+2)=840 \end{align} Let $y=n^2-3 n$ then the above last equati... \Rightarrow \text{Either} y&=28, \text{or} y=-30\end{align} When $y=28$ then \begin{align} & n^2-3 n=2
- Question 5 Exercise 6.1
- n-4))\\ &(2 n-(2 n-3))(2 n-(2 n-2))(2 n-(2 n-1))]\end{align} In the L.H.S of the above equation are tot... ots 4.2] {[(2 n-1)(2 n-3)(2 n-5) \ldots 5.3 .1]} \end{align} Each underlined brackets contain $n$ terms... \ \dfrac{(2 n) !}{n !}&=2^n(1.3 .5 \ldots(2 n-1))\end{align} =====Question 5(ii)===== Show that: $\df... n-2))(2 n+1-(2 n-1))(2 n+1-(2 n-2))(2 n+1-(2 n))]\end{align} In the L.H.S of the above equation. are to
- Question 1 Exercise 6.4
- {5\}\\ P(A)&=\dfrac{n(A)}{n(S)}\\ &=\dfrac{1}{6} \end{align} =====Question 1(b)===== Let $S=\{1,2,3,4,... \ P(B)&=\dfrac{n(B)}{n(S)}\\ &=\dfrac{0}{6}\\ &=0\end{align} =====Question 1(c)===== Let $S=\{1,2,3,4,... \ P(C)&=\dfrac{n(C)}{n(S)}\\ &=\dfrac{6}{6}\\ &=1\end{align} =====Question 1(d)===== Let $S=\{1,2,3,4,... rac{n(D)}{n(S)}\\ &=\dfrac{2}{6}\\ &=\dfrac{1}{3}\end{align} =====Question 1(e)===== Let $S=\{1,2,3,4,
- Question 3 and 4 Exercise 6.5
- htarrow P(B)&=P(A \cup B)-P(A)\\ &=0.6-.0 .5=0.1 \end{align} =====Question 4===== A bag contains $30$ ... S&=\{1,2,3, \ldots, 50\} \text { so }\\ n(S)&=50 \end{align} Let \begin{align}A \{odd \,numbers \}&=\{1... f\, an\, integer \}\\ &=\{1,4,9,16,25\}\\ n(B)&=5\end{align} Álso \begin{align}A \cap B&=\{1,9,25\}\\ \text{Therefore} n(A \cap B)&=3\end{align} Now $$P(A)=\dfrac{15}{30}, P(B)=\dfrac{5}{
- Question 3 & 4 Exercise 6.1
- !} \\ & =\dfrac{56+16+3}{8 !}\\ &=\dfrac{75}{8 !}\end{align} =====Question 3(ii)===== Prove that $\dfr... (n+3) !}{(n+3) !} \\ & =(n+5)(n+4)\\ &=n^2+9 n+20\end{align} =====Question 4(i)===== Find the value of... ghtarrow \text { Either. } n&=-2 \text { or } n=6\end{align} $n$ can not b negative, therefore $n=6$. ... \dfrac{n(n-1) !}{(n-1) !}&=9 \\ \Rightarrow n&=9 \end{align} ====Go To==== <text align="left"><b
- Question 10 Exercise 6.2
- \cdot 6 \cdot 5 \cdot 4 \cdot 3 !}{3 !}\\ &=6720\end{align} If certain two students insist to sit next... \\ &=2 \times\dfrac{7.6 .5 .4 .3 !}{3 !}\\ &=1680\end{align} =====Question 10(ii)===== In how many way... \cdot 6 \cdot 5 \cdot 4 \cdot 3 !}{3 !}\\ &=6720\end{align} If certain two students refuse to sit next... !}{3 !}\\ &=840\\ \text{then}\quad &6720-840=5880\end{align} ====Go To==== <text align="left"><b