Contains the center? 2

The solution is identical to the one I posted for Contains the Center 1 .

Call the given points $P_1, \dots, P_5$ .

In order for the center of the circle not to lie in the quadrilateral, there must be an arc of more than $180^\circ$ without points. Let $P_a$ be the point immediately after this long arc in clockwise direction. The other four points then lie within $180^\circ$ , or half a circle, clockwise from $P_a$ . The probability that they do so is $\tfrac12$ for each point, or $(\tfrac12)^4 = \tfrac1{16}$ combined.

That first point $P_a$ may be any of the five chosen points. Therefore the total probability that the center of the circle does not lie in the quadrilateral is $5 \times \tfrac1{16} = \tfrac5{16}$ .

The probability that the center of the circle does lie inside the quadrilateral is the complement, $1 - \tfrac{11}{16} = \tfrac7{16} \approx \boxed{0.6875}$ .