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Question 1 Exercise 6.3
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on==== We are given: \begin{align}&^n C_2=36\\ & \Rightarrow \dfrac{n !}{(n-2) ! 2 !}=36 \\ & \Rightarrow \dfrac{n(n-1)(n-2) !}{(n-2) ! \cdot 2}=36 \\ & \Rightarrow n(n-1)=72 \\ & \Rightarrow n^2-n-72=0 \\ & \Rightarrow n^2-9 n+8 n-72=0\\ & \Rightarrow n(n-9)+8(n-9)=0 \\ &
Question 1 and 2 Exercise 6.2
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We are given: \begin{align}^n P_5&=56(^n P_3) \\ \Rightarrow \dfrac{n !}{(n-5) !}&=56 \dfrac{n !}{(n-3) !} \\ \Rightarrow \dfrac{1}{(n-5) !}&=\dfrac{56}{(n-3) !} \\ \Rightarrow \dfrac{1}{(n-5) !}&=\dfrac{56}{(n-3)(n-4)(n-5) !} \\ \Rightarrow(n-3)(n-4)&=56 \\ \Rightarrow n^2-7 n+12-56&=0 \\
Question 2 Review Exercise 6
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begin{align} { }^{2 n} C_r&={ }^{2 n} C_{r+2} \\ \Rightarrow \dfrac{(2 n) !}{(2 n-r) ! r !}&=\dfrac{(2 n) !}{(... ing both sides by $(2 n)$ ! we get \begin{align} \Rightarrow \dfrac{1}{(2 n-r) ! r !}&=\dfrac{1}{(2 n-r-2) !(r+2) !} \\ \Rightarrow \dfrac{1}{(2 n-r)(2 n-r-1)(2 n-r-2) ! r !}&=\dfrac{1}{(2 n-r-2) !(r+2) !} \\ \Rightarrow \dfrac{1}{(2 n-r)(2 n-r-1) r !}& =\dfrac{1}{(r+2)
Question 2 Exercise 6.3
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gn} & { }^n C_4=\dfrac{n !}{(n-4) ! 4 !}=35 \\ & \Rightarrow \dfrac{n(n-1)(n-2)(n-3)(n-4) !}{(n-4) !}=35 \times 24=840 \\ & \Rightarrow n(n-1)(n-2)(n-3)=840 \\ & \Rightarrow n(n-3)(n-1)(n-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
Question 3 & 4 Exercise 6.1
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ac{n(n !)}{(n-5) !}&=\dfrac{12(n !)}{(n-4) !} \\ \Rightarrow \dfrac{n}{(n-5) !}&=\dfrac{12}{(n-4)(n-5) !} \\ \Rightarrow n&=\dfrac{12}{(n-4)} \\ \Rightarrow n(n-4)&=12 \\ \rightarrow n^2-4 n-12&=0 \\ \rightarrow n^2+2 n-6 n \quad 12&=0 \\ \Rightarrow n(n+2)-6(n+2)&=
Question 9 Exercise 6.5
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n} 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}\en... text { Both are selected })&=P(A) \times P(B) \\ \Rightarrow P(\text { Both are selected }) & =\dfrac{1}{7} \t... n} P(\text { Ajmal scicction })&=\dfrac{1}{7} \\ \Rightarrow P(\text { Ajmal not selected })&=\dfrac{6}{7} \\
Question 1 and 2 Exercise 6.5
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n{align} P(A \cup B)&=P(A)+P(B)-P(A \cap B) \\ \Rightarrow P(A \cap B)&=P(A)+P(B)-P(A \cup B) \end{align} Su... 1}{2}\\ P(\bar{A} \cap \bar{B})&=1-P(A \cup B)\\ \Rightarrow P(\bar{A} \cap \bar{B})&=1-\dfrac{1}{2}=\dfrac{1}... egin{align}P(A \cap B)&=P(A)+P(B)-P(A \cap B) \\ \Rightarrow P(A \cap B)&=P(A)+P(B)-P(A \cup B)\end{align} Usi... ap B)&=\dfrac{1}{2}+\dfrac{3}{8}-\dfrac{3}{4} \\ \Rightarrow \quad P(A \cap B)&=\dfrac{4+3-6}{8} \\ \Rightarro
Question 3 Exercise 6.3
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in{align} & { }^{2 n} C_3:{ }^n C_2=36: 3 . \\ & \Rightarrow \dfrac{(2 n) !}{(2 n-3) ! 3 !} \times \dfrac{(n-2) ! 2 !}{n !}=12 \\ & \Rightarrow \dfrac{2 n(2 n-1)(2 n-2)(2 n-3) !}{(2 n-3) ! 3 !}... \times\dfrac{(n-2) ! 2 !}{n(n-1)(n-2) !}=12 \\ & \Rightarrow \dfrac{4 n(2 n-1)(n-1)}{3} \cdot \dfrac{1}{n(n-1)}=12 \\ & \Rightarrow 2 n-1=9 \\ & \Rightarrow 2 n=10 \\ & \Rightarrow
Question 3 & 4 Review Exercise 6
6 Hits, Last modified:
n} { }^{56} P_{r+6}:{ }^{54} P_r+3&=30800: 1 \\ \Rightarrow \dfrac{\dfrac{56 !}{[56-(r+6)] !}}{\dfrac{54 !}{[54-(r+3)] !}}&=\dfrac{30800}{1} \\ \Rightarrow \dfrac{56 !}{(50-r) !} \times \dfrac{(51-r) !}{54 !}&=30800 \\ \Rightarrow \dfrac{56.55 .54 !}{(50-r) !} \times \dfrac{(51-r)(50-r) !}{54 !}& =30800 \\ \Rightarrow \dfrac{3080}{1} \times \dfrac{51-r}{1}&=30800\\ \
Question 10 Exercise 6.5
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gin{align} P(A \cup B)&=P(A)+P(B)-P(A \cap B) \\ \Rightarrow P(A \cup B)&=\dfrac{190}{435}+\dfrac{231}{435}-\dfrac{105}{435} \\ \Rightarrow P(A \cup B)&=\dfrac{190+231-105}{435} \\ \Rightarrow P(A \cup B)&=\dfrac{316}{435}\end{align} ====Go To===
Question 7 & 8 Review Exercise 6
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align} P(B \mid A)&=\dfrac{P(A \cap B)}{P(A)} \\ \Rightarrow P(A \cap B)&=P(B \mid A) \cdot P(A)\\ &=0.4 \time... gin{align} P(A \cup B)&=P(A)+P(B)-P(A \cap B) \\ \Rightarrow P(A \cup B)&=0.8+0.5-0.32=0.98\end{align} =====Q
Question 4 Exercise 6.3
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{align} r^n C_r&=r \dfrac{n !}{(n-r) ! r !} \\ \Rightarrow r^n C_r&=r \dfrac{n(n-1) !}{(n-1-(r-1)) r(r-1) !}
Question 3 and 4 Exercise 6.5
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et$. Thus \begin{align}P(A \cup B)&=P(A)+P(B)\\ \Rightarrow P(B)&=P(A \cup B)-P(A)\\ &=0.6-.0 .5=0.1 \end{ali
Question 11 Review Exercise 6
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P(\text { Red })+P( Green )-P( Red and Green )\\ \Rightarrow P(Red or Green )&=\dfrac{1}{4}+\dfrac{1}{4}-\phi\