The middle number

If $0 < x < 1$ and $\delta x$ is an infinitesimal, the probability that the middle number $X$ lies between $x$ and $x + \delta x$ is $P[x \le x \le x + \delta x] \; = \; 5 \times \binom{4}{2} \times x^2 \times (1-x - \delta x)^2 \times \delta x\; \approx \; 30x^2(1-x)^2\delta x$ (two of the numbers must be less than $x$ , two must be greater than $x + \delta x$ , one must be between $x$ and $x + \delta x$ , and there are $5 \times \binom{4}{2} = 30$ ways in which this can happen). Thus the probability density function of $X$ is $f_X(x) \; = \; \left\{ \begin{array}{lll} 30x^2(1-x)^2 & \hspace{1cm} & 0 < x < 1 \\ 0 & & \mathrm{o.w.}\end{array}\right.$ and hence the desired probability is $P[\tfrac25 < X< \tfrac35] \; = \; 30\int_{\frac25}^{\frac35} x^2(1-x)^2\,dx \; = \; \tfrac{1141}{3125}$ making the answer $1141 + 3125 = \boxed{4266}$ .

Calculate the probability $P(m,n,k)$ -here $m$ - numbers are between $0$ and $0.4$ , $n$ - numbers are between $0.4$ and $0.6$ , $k$ - numbers are between $0.4$ and $0.6$ , $m+n+k=5$ .

$P(2,1,2)=C^2_{5}C^1_{3} {\frac{2^4}{5^4}} \frac{1}{5}=\frac{480}{5^5}$

$P(1,3,1)=C^1_{5}C^1_{4} {\frac{2^2}{5^2}} \frac{1}{5^3}=\frac{80}{5^5}$

$P(2,2,1)=P(1,2,2)=C^1_{5}C^2_{4} {\frac{2^3}{5^3}} \frac{1}{5^2}=\frac{240}{5^5}$

$P(0,3,2)=P(2,3,0)=C^1_{5} {\frac{2^2}{5^2}} \frac{1}{5^3}=\frac{40}{5^5}$

$P(0,4,1)=P(1,4,0)=C^1_{5} {\frac{2}{5}} \frac{1}{5^4}=\frac{10}{5^5}$

$P(0,5,0)= \frac{1}{5^5}$

$\sum {P(m,n,k)}= \frac{480}{5^5}+\frac{80}{5^5}+2 \cdot\frac{240}{5^5}+2\cdot\frac{40}{5^5}+2\cdot\frac{10}{5^5} +\frac{1}{5^5}=\frac{1141}{3125}$