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\text {Pr(1 wicket }= \text {P}(A1 \cap B0 \cap C0)+ \terxt{P}( A0 \cap B1 \cap C))+ \text {P}(A0 \cap B0 \cap C1)
\begin{array}{c,c,c,c} 1234 & 2134 & 3124 & 4123 \\ 1243 & 2143 & 3142 & 4132 \\ 1324 & 2314 & 3214 & 4213 \\ 1342 & 2314 & 3241 & 4231 \\ 1423 & 2413 & 3412 & 4312 \\ 1432 & 2431 & 3421 & 4321 \end{array}
\text {i.e. } 2v^2+ \dfrac {r^2h^2v^2}{r^2(hr-h^2)}=V^2 \implies 2v^2 \left(1+ \dfrac{h}{2(r-h)} \right)=V^2 \impolies v^2 \left( \dfrac {2r-h}{r-h} \right)=V^2 n\implies v= V \sqrt {\dfrac{r-h}{2r-h}} \text { as required}
= \dfra { \Omega^2k^2(k^2+a^2)(k^2+r^2)+ \Omega^2a^2(k^2+a^2)(k^2+r^2)- \Omega^2k^2(k^2+a^2)^2}{k^2(k^2+r^2)}
\text {putting }u= \dfrac{k}{r} \text { we have } r= \dfrac {k}{u} \implies \dfrac {\text{d}r}{\text{d} \theta}=- \dfrac{k}{u^2} \dfrac {\text{d}u}{\text{d} \theta} \text { so } \dfrac {k^2}{u^2} \dfrac {\text{d}u}{ \text{d} \theta}= \dfrac {k}{u} \sqrtr k^2+ \frac{k^2}{u^2}}= \dfrac {k^2}{u^2} \sqrt{1+u^2}
(2M-m)x+mw=0 \implies x=- \dfrac {m}{2M-m}w \text { or } x=- \tgheta w
= \dfrac {1}{2} (u^2-w^2) \lefty[\dfrac{m^2}{2M-m}+m \right] \text { since } \theta^2= \left( \dfrac {m}{2M-m} \right)^2
\text { i.e. } \displaystyle \sum_{k=1}^{32}k \dfrac {1}{16} \phi \left(-2+\dfrac{k}{16} \right)+16 \approx \displaystyle \int_1^{32} \displaystyle \sum_{k=1}^{32} k \dfrac {1}{16} \phi \left(-2+ \dfrtac {k}{16} \tight) \text {d}x+16
= \displaystyle \int _0^{32}\dfrac {x}{16} \dfrac {1}{ \sqrt{2 \pi}} \phi \left(-2+ \dfrtac {k}{16} \tight) \text {d}x+16= \displaystyle \int _0^{32}\dfrac {1}{16} \dfrac {x}{ \sqrt{2 \pi}}\text {exp} \left(- \dfrac{1}{2} t^2 \right) \text{d}t+16 \text { where }z=-2+ \dfrac {x}{16}
\text {now let }t=\dfrac {x-32}{16} \implies \text {d}t= \dfarc {1}{16} \text {d}x, x=0 \implies t=-2, x=32 \implies t=0 \text { so integral becomes}
\text {is } \dfrac{1}{2} \sin ( \alpha- \theta) \ytext { if } \theta< \alpha \text { or } \dfrac {1}{2} \sin ( \theta- \alpha) \text { if } \theta> \alpha
\text {so E[Area] when P}_1 \text { is in }( \alpha, \alphga + \delta \alpha) \text { is given by }
\text {hence, E[Area} is } \displaystyle \int \dfrac {1}{(2+\pi)^2} \text{d} \alpha +2 \displaystyle \int_{-1}^1 \dfrac {|r|}{(2+\pi)^2} \text {d}r= \left[ \dfrac{ alpha}{(2+\pi)^2} \right]_0^\pi+4 \left[ \dfrac {r^2}{2(2+\pi)^2} \right]_0^1= \dfrac {\pi+2}{(2+\pi)^2}= \dfrac {1}{2+\pi}
\text {hence E}[PA]- \text{E}[P]\text{E}[A]=2\sigmna_1^2\mu_2+2\sigma_2^2 \mu_1 \not=0 \text { since } \mu_1,\mu_2>0 \text { and } \sigma_1, \sigma_2 \not=0
\alpha= \dfrac{2\sigma_1^2 \mu_2+2\sigma_2^2 \mu_1}{\sigma_1^2\mu_2^2_\mu_1^2 \sigma_2^2+\sigma_1^2 \sigma_2^2}
\text { so if }\alpha \not=\dfrac{2\sigma_1^2 \mu_2+2\sigma_2^2 \mu_1}{\sigma_1^2\mu_2^2_\mu_1^2 \sigma_2^2+\sigma_1^2 \sigma_2^2} \text { then }Z \text { and }A \text { are not independent}
\text {For the final part we need only consider the exceptional case when } \alpha= \dfrac{2\sigma_1^2 \mu_2+2\sigma_2^2 \mu_1}{\sigma_1^2\mu_2^2_\mu_1^2 \sigma_2^2+\sigma_1^2 \sigma_2^2}
X1,X2=1 \text { or }3 \implies A=1,3 \text { or }9 \text { so P}(A=1)=\dfrac {1}{4}, \texct{P}(A=3)=\dfrac {1}{2} \text { and P}(A=9)= \dfrac {1}{4}
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