# How to calculate this probability?

Here I want to find the probability $p$ which is given by

$$p=\int_0^{r_d}\mathcal{L}_I(\beta r^{\alpha}|r)f_R(r)dr$$

with $r_d=1$ and $0 \le r\le r_d$.

I want to calulate the probability for different values of $\beta$ and $\alpha$

Here,

$f_R(r)=2Nr(1-r^2)^{N-1}$

$\mathcal{L}(\beta r^{\alpha}|r)=\left(\frac{\mathcal{C}(\alpha,\beta r^{\alpha},r_d)-\mathcal{C}(\alpha,\beta r^{\alpha},r)}{r_d^2-r^2}\right)^{p_bN-1}$

$\mathcal{C}(\alpha,\beta r^{\alpha},x)=x^2-x^2\hspace{1mm} _2F_1\left(1,2/\alpha,1+2/\alpha,-x^{\alpha}/(\beta r^{\alpha})\right)$

rd=1;
N=5;
pb=1;
c[α_,r_,x_]:=x^2-x^2*Hypergeometric2F1[1,2/[Alpha],1+2/[Alpha],-x^α/(β*r^α)];

β=3.1623;
α=4;

p=Integrate[((c[α_,r,rd]-c[α_,r,r])/(rd^2-r^2))^(N-1)*2*N*r*(1-r^2)^(N-1),{r,0,1}]


I am getting some syntax error.

Do you mean

rd = 1;
n = 5;
pb = 1;
c[α_, r_, x_] := x^2 - x^2*Hypergeometric2F1[1, 2/α, 1 + 2/α, -x^α/(β*r^α)];

β = 3.1623;
α = 4;

p = Integrate[((c[α, r, rd] - c[α, r, r])/(rd^2 - r^2))^(n - 1)*2*n*r*(1 - r^2)^(n - 1),
{r, 0, 1}]

Integrate[10*r*(1 - 0.08904220055024259*r^2 -
Hypergeometric2F1[1/2, 1, 3/2, -(0.31622553204945764/r^4)])^4, {r, 0, 1}]

N@%


0.416275

Avoid using capital letters as variables so that they don't conflict with Mathematica's existing definitions/functions (here "N" was part of the problem). There were some erroneous underscores as well, used when referencing a function from another function.

rd = 1;
n = 5;
pb = 1;
c[α_, r_, x_] := x^2 - x^2*Hypergeometric2F1[1, 2/ α, 1 + 2/ α, -x^α/(β*r^α)];

β = 3.1623;
α = 4;

p = Integrate[((c[α, r, rd] - c[α, r, r])/(rd^2 - r^2))^(n - 1)*2*n*r*(1 - r^2)^(n - 1),
{r, 0, rd}]


Integrate[10*r*(1 - 0.08904220055024259*r^2 - Hypergeometric2F1[1/2, 1, 3/2, -(0.31622553204945764/r^4)])^4, {r, 0, 1}]

N@%


0.416275

or should

p = Integrate[((c[α, β*r^α, rd] - c[α, β*r^α, r])/(rd^2 - r^2))^(n - 1)*2*n*
r*(1 - r^2)^(n - 1), {r, 0, rd}]