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- Question 4 Exercise 6.4
- ossed three times. Hence the sample space of the given problem is: \begin{align}S&=(HHII,HHT.HTH.HTT.THI... ossed three times. Hence the sample space of the given problem is: \begin{align}S&=(HHII,HHT.HTH.HTT.THI... ossed three times. Hence the sample space of the given problem is: \begin{align}S&=(HHII,HHT.HTH.HTT.THI... ossed three times. Hence the sample space of the given problem is: \begin{align}S&=(HHII,HHT.HTH.HTT.THI
- Question 3 & 4 Exercise 6.1
- Solution==== We are taking the L.H.S of the above given equation. \begin{align}\dfrac{1}{6 !}+\dfrac{2}{7... Solution==== We are taking the L.H.S of the above given equation. \begin{align}\dfrac{(n+5) !}{(n+3) !}&=... \dfrac{12(n !)}{(n-4) !}$ ====Solution==== We are given: \begin{align}\dfrac{n(n !)}{(n-5) !}&=\dfrac{12(... c{(n-1) !}{(n-4) !}=9: 1$ ====Solution==== We are given: \begin{align} \dfrac{n !}{(n-4) !}: \dfrac{(n-1)
- Question 1 and 2 Exercise 6.5
- }{4}$. Find $P(A \cap B)$ ====Solution==== We are given: $$P(A)=\dfrac{1}{2}, P(\bar{B})=\dfrac{5}{8}, P(... =P(A)+P(B)-P(A \cup B)\end{align} Using the above given and calculated \begin{align} P(A \cap B)&=\dfrac{... P(\bar{A} \cap \bar{B})$. ====Solution==== We are given: $$P(A)=\dfrac{1}{2}, P(\bar{B})=\dfrac{5}{8}, P(... })-P(\bar{A} \cup \bar{B})\end{align} Putting the given and calculated \begin{align}P(\bar{A} \cap \bar{B
- Question 1 and 2 Exercise 6.2
- P_5=56(^n P_3)$ for $n.$ ====Solution==== We are given: \begin{align}^n P_5&=56(^n P_3) \\ \Rightarrow \... 5=9(^{n-1} P_4)$ for $n.$ ====Solution==== We are given: \begin{align} ^n P_5&=9(^{n-1} P_4) \\ \Rightarr... lve $n^2 P_2=600$ for $n$ ====Solution==== We are given: \begin{align}n^2 P_2&=60 \\ \Rightarrow \dfrac{(
- Question 1 Exercise 6.3
- lve $^n C_2=36$ for $n$. ====Solution==== We are given: \begin{align}&^n C_2=36\\ & \Rightarrow \dfrac{n... _4=6,^{n-1} C_2$ for $n$. ====Solution==== We are given: \begin{align} & { }^{n+1} C_4=6 .^{n-1} C_2 \\ &... 2=30 .{ }^n C_3$ for $n$. ====Solution==== We are given: \begin{align} & { }^2 C_2=30 .^n C_3 \\ & \Right
- Question 3 and 4 Exercise 6.2
- ^n P_r=n(^{n-1} P_{r-1})$ ====Solution==== We are given that: $$^n P_r=n({ }^{n-1} P_{r-1})$$ We are taki... 1} P_r+r(^{n-1} P_{r-1})$ ====Solution==== We are given: $$^n P_r=^{n-1} P_r+r({ }^{n-1} P_{r-1})$$ Takin
- Question 4 Exercise 6.3
- C_r+{ }^{n-1} C_{r-1}={ }^n C_r$ ====Solution==== Given that: $${ }^n{ }^1 C_r+{ }^n{ }^1 C_{r-1}={ }^n C... \cdot{ }^n C_r=n^{n-1} C_{r-1}$ ====Solution==== Given that: $$r^n C_r=n^{n-1} C_{r-1} $$ We take L.H.S
- Question 3 and 4 Exercise 6.5
- KPTBB) Peshawar, Pakistan. =====Question 3===== Given $P(A)=0.5$ and $P(A \cup B)=0.6$. find $P(B)$ if ... $ are mutually exclusive. ====Solution==== We are given that $\mathrm{A}$ and $B$ are mutually exclusive,
- Question 5 and 6 Exercise 6.5
- $E^{\prime}$ are complementary events, and we are given $P(E)=\dfrac{8}{9}$, therefore, \begin{align}P(E^... =====Question 6===== In the two dice experiment, given that the the first dice shows $4$, what is the pr
- Question 2 Review Exercise 6
- _r={ }^{2 n} C_{r+2}$; find $r$. ====Solution==== Given that: \begin{align} { }^{2 n} C_r&={ }^{2 n} C_{r... {18} C_{r+2}$; find ${ }^r C_5$. ====Solution==== Given that: \begin{align} { }^{18} C_r&={ }^{18} C_{r+2
- Question 9 Exercise 6.2
- an. =====Question 9===== How many signals can be given by six flags of different colors when any number
- Question 2 Exercise 6.3
- =840$ and ${ }^n C_r=35$. ====Solution==== We are given: \begin{align} &^n P_r=\dfrac{n !}{(n-r) !}=840 .
- Question 5 and 6 Exercise 6.3
- answer is "No," because there are no three of the given points on any line. =====Question 5(ii)===== How
- Question 3 & 4 Review Exercise 6
- 54} P_{r+3}=30800: 1$. Find $r$. ====Solution==== Given that: \begin{align} { }^{56} P_{r+6}:{ }^{54} P_r
- Question 7 & 8 Review Exercise 6
- the total telephone that can be formed with the given digits and each one starts with $25$ are: $$m_1 \