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- Question 9 Exercise 6.5
- ll be selected. ====Solution==== \begin{align} P(\text { Ajmal scicction })&=\dfrac{1}{7} \\ \Rightarrow P(\text { Ajmal not selected })&=\dfrac{6}{7} \\ P(\text { Bushra selection })&=\dfrac{1}{5} \\ \Rightarrow P(\text { Bushra not selected })&=\dfrac{4}{5}\end{align}
- Question 13 Exercise 6.2
- $ and $m_3=2$ are $C$. Therefore, \begin{align}\text{total number of permutations are} &=\left(\begin... L$ and $m_3=2$ are $C$. Therefore, \begin{align}\text{Number of permulations are} &=\left(\begin{array}... L$ $m_3=2$ are $C$. Therefore, \begin{align}\text{total number of permutations are} & =\left(\begin... and $m_3=1$ are $C$. $\therefore$ \begin{align}\text{Number of permutations are}& =\left(\begin{array}
- Question 2 Exercise 6.3
- ac{(n-r) ! r !}{n !}&=\dfrac{840}{35}\\ r!&=24\\ \text{or}\quad r &=4\end{align} Putting $r=4$ in Eq.(ii... 30)=0\\ &\Rightarrow(y-28)(y+30)=0\\ \Rightarrow \text{Either} y&=28, \text{or} y=-30\end{align} When $y=28$ then \begin{align} & n^2-3 n=28 \\ & \Rightarrow ... $n$ can not be negative, so $n=7$. \begin{align}\text{When} &y=-30 \text{then}\\ & n^2-3 n=-30 \\ & \Ri
- Question 1 Exercise 6.3
- =0 \\ & \Rightarrow(n-9)(n+8)=0 \\ & \Rightarrow \text { either } n=9 \text { or } n=-8\end{align} But $n$ can not be negative therefore, we have $n=9$. ====... =0 \\ & \Rightarrow(n+9)(n-8)=0 \\ & \Rightarrow \text { either } n=-9 \text { or } n=8 \end{align} But $n$ can not be negative, therefore we have $n=8$. ==
- Question 3 and 4 Exercise 6.5
- ickets 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)&=15\\ \text{Let}\, B&=\{ number \,square \,of\, an\, integer ... {align} Álso \begin{align}A \cap B&=\{1,9,25\}\\ \text{Therefore} n(A \cap B)&=3\end{align} Now $$P(A)=\... }=\dfrac{17}{30}\end{align} ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:so
- Question 11 Review Exercise 6
- e \begin{align}\boldsymbol{P}( Red or Green )&=P(\text { Red })+P( Green )-P( Red and Green )\\ \Rightar... hbf{P}( Not Red)$ The probability of red is: $$P(\text { Red })=\dfrac{1}{4}$$ Then by complementary event theorem: \begin{align} P(\text { not red })&=1-P(\text { Red }) \\ & =1-\dfrac{1}{4}=\dfrac{3}{4}\end{align} $P(Pink)$ Since pink co
- Question 3 & 4 Exercise 6.1
- 2)&=0 \\ \Rightarrow(n-6)(n+2)&=0 \\ \Rightarrow \text { Either. } n&=-2 \text { or } n=6\end{align} $n$ can not b negative, therefore $n=6$. =====Question 4... Rightarrow n&=9 \end{align} ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit06:ex6-1-p1 |< Question 1 & 2 ]]</btn></text> <text align="right"><btn type="success">[[math-1
- Question 1 and 2 Exercise 6.2
- -11(n+4)&=0 \\ \Rightarrow(n-11)(n+4)&=0\\ n=11& \text{or}\quad n=-4\end{align} But $n$ can not be negat... \\ \Rightarrow(n^2-25)(n^2+24)=0\\ n^2-25=0\quad\text{or}\quad n^2+24&=0\\ n^2=25\quad\text{or}\quad n^2&=-24\\ n=\pm 5\quad\text{or}\quad n&=\sqrt{24}i\end{align} but $n$ can not be negative nor
- Question 1 Exercise 6.4
- less than $1$ Let \begin{align}B&=\{\}\\ &=\phi \text{then}\\ P(B)&=\dfrac{n(B)}{n(S)}\\ &=\dfrac{0}{6}... ter than $0$ Let \begin{align}C&=\{1,2,3,4,5,6\},\text{then}\\ P(C)&=\dfrac{n(C)}{n(S)}\\ &=\dfrac{6}{6}... ing a multiple of $3$ Let \begin{align}D&=\{3,6\}\text{then}\\ P(D)&=\dfrac{n(D)}{n(S)}\\ &=\dfrac{2}{6}... n or equal to $4$ Let \begin{align} E &=\{4,5,6\}\text{then}\\ P(E)&=\dfrac{n(E)}{n(S)}\\ &=\dfrac{3}{6}
- Question 4 Exercise 6.4
- {align}S&=(HHII,HHT.HTH.HTT.THII.THT.TTH,TT7),\\ \text{then} n(S)&=2^3=8\end{align} When all heads. Let... align}S&=(HHII,HHT.HTH.HTT.THII.THT.TTH,TT7),\\ \text{then} n(S)&=2^3=8\end{align} When two heads Let ... ac{n(F)}{n(S)}=\dfrac{1}{8}$$ ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit06:ex6-4-p3 |< Question 3 ]]</btn></text> <text align="right"><btn type="success">[[math-1
- Question 5 and 6 Exercise 6.2
- though these 5 digits are: $$=5.4 .3 .2=120\quad \text{or}$$ can be found using permutation as: $$^5 P... dot m_4=2 \cdot 3 \cdot 2=12$$ ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit06:ex6-2-p2 |< Question 3 & 4 ]]</btn></text> <text align="right"><btn type="success">[[math-11-kpk:sol:unit06:ex6-2-p4|Question 7 & 8 >]]</btn></te
- Question 10 Exercise 6.2
- -4) !}\\ &=\dfrac{7.6 .5 .4 .3 !}{3 !}\\ &=840\\ \text{then}\quad &6720-840=5880\end{align} ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit06:ex6-2-p5 |< Question 9 ]]</btn></text> <text align="right"><btn type="success">[[math-11-kpk:sol:unit06:ex6-2-p7|Question 11 >]]</btn></text>
- Question 3 Exercise 6.4
- C_6+{ }^8 C_7+{ }^8 C_8 \\ & =28+8+1=37 . \quad \text { Thus } \\ P(D)-\dfrac{n(D)}{n(D)}&=\dfrac{37}{256}\end{align} ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit06:ex6-4-p2 |< Question 2 ]]</btn></text> <text align="right"><btn type="success">[[math-11-kpk:sol:unit06:ex6-4-p4|Question 4 >]]</btn></text>
- Question 5 and 6 Exercise 6.5
- \{ a\, number\, greater\, tha, 4\}\\ &=\{5,6\}\\ \text{then} n(A)&=2\end{align} Thus, the probability th... 6}\\ &=\dfrac{1}{3}\end{align} ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit06:ex6-5-p2 |< Question 3 & 4 ]]</btn></text> <text align="right"><btn type="success">[[math-11-kpk:sol:unit06:ex6-5-p4|Question 7 >]]</btn></text>
- Question 2 Review Exercise 6
- r&=304 \\ \Rightarrow r&=\dfrac{304}{38}=8 . \\ \text { Now }{ }^r C_5&={ }^8 C_5=\dfrac{8 !}{(8-5) ! 5 !}=56\end{align} ====Go To==== <text align="left"><btn type="primary">[[math-11-kpk:sol:unit06:Re-ex6-p1 |< Question 1 ]]</btn></text> <text align="right"><btn type="success">[[math-11-kpk:sol:unit06:Re-ex6-p3|Question 3 & 4 >]]</btn></t