Probability

3, 4, 5, 5

In remaining six places you have to arrange

3, 4, 5,5

So no. of ways $=\frac{6!}{2!2!2!}$

Total no. of seven digits nos. = $\frac{7!}{2!3!2!}*1$

Hence Req. prob. $\frac{\frac{6!}{2!2!2!}}{\frac{7!}{2!3!2!}}=\frac{6!3!}{2!7!}=\frac{3}{7}$

Let n be the number of times.

$p=\frac{1}{2},q=\frac{1}{2}$

According to question,

${\Rightarrow }^n{C}_7{=}^n{C}_9\Rightarrow n-7=9\Rightarrow n=16$

$p\left(x=2\right){=}^{16}{C}_2{\left(\frac{1}{2}\right)}^2.{\left(\frac{1}{2}\right)}^{14}=\frac{15}{2^{13}}$

Let total number of throws = n

Probability of getting 2 times = Probability of getting an even number 3 times.

[as probability of getting odd number = probability of getting even number = $\frac{1}{2}$ ]

Probability of getting an odd number for odd number of times =

Let P (B₁) = a      P (B₂) = b            P (B₃) = c

Given a (1 – b) (1 – c) = a . (i)

b (1 – a) (1 – c) = b              . (ii)

c (1 – b) (1 – a) = $\gamma$             . (iii)

(1 – a) (1 – b) (1 – c) = p     . (iv)

$\left(\alpha -2\beta \right)p=\alpha \beta$  

->a – ab – 2b + 2ab = ab Þ a = 2b . (v)

Again $\left(\beta -3\gamma \right)p=2\beta \gamma$  

->b – bc – 3c + 3bc = 2bc Þ b = 3c   . (vi)

$\Rightarrow \frac{P\left(B_1\right)}{P\left(B_3\right)}=\frac{a}{c}=\frac{2b}{b/3}=6$      

  $z+\alpha \left|z-1\right|+2i=0$

Let z = x + iy

$\Rightarrow x+i\left(y+2\right)=-\alpha \sqrt[]{{\left(x-1\right)}^2+y^2}$            

 for $\alpha \in Ry+2=0\Rightarrow y=-2$  

$x^2={\alpha }^2\left[{\left(x-1\right)}^2+4\right]$      = 0

$\frac{1}{{\alpha }^2}\ge \frac{4}{5}\Rightarrow {\alpha }^2\le \frac{5}{4}\Rightarrow \frac{-\sqrt[]{5}}{2}\le \alpha \le \frac{\sqrt[]{5}}{2}$

$4\left(p^2+q^2\right)=4\left(\frac{5}{4}+\frac{5}{4}\right)=10$

  $\frac{2^{21}{\left(\frac{-1+i\sqrt[]{3}}{2}\right)}^{21}}{{\left(\sqrt[]{2}\right)}^{24}{\left(\frac{1-i}{\sqrt[]{2}}\right)}^{24}}+\frac{2^{21}{\left(\frac{1+\sqrt[]{3}i}{2}\right)}^{21}}{{\left(\sqrt[]{2}\right)}^{24}{\left(\frac{1+i}{\sqrt[]{2}}\right)}^{24}}=k$

=   ${\displaystyle \sum _{j=0}^5\left(j+5\right)\left(j+5-1\right)={\displaystyle \sum _{j=0}^5\left(j+5\right)\left(j+4\right)}}$

= ${\displaystyle \sum _{j=0}^5\left(j^2+9\widehat{j}+20\right)=\frac{5*6*11}{6}+\frac{9*5*6}{2}+20*6}$  

55 + 135 + 120 = 310

Total possibilities = 2⁵ * 2⁵

Farounable case = ⁵C₂ * 3³ = 10 * 3³

$\therefore requiredprobability=\frac{10*3^3}{2^5*2^5}=\frac{5*27}{2^9}=\frac{135}{2^9}$

P {a person chosen from the group has chest disorder}

$=\frac{160}{400}*0.35+\frac{100}{400}*0.20+\frac{140}{400}*0.10$

$P\left(\frac{Personissmoker\&non-vegetarian}{Personhaschestdisorder}\right)$

$=\frac{\frac{160}{400}*0.35}{\frac{160}{400}*0.35+\frac{100}{400}*0.20+\frac{140}{400}*0.10}$

$=\frac{40*0.7}{40*0.7+25*0.4+35*0.2}=\frac{23}{23+10+7}=\frac{28}{45}$