ABCD is a rhombus with diagonals $$AC = 10\ \text{cm}$$ and $$BD = 10\sqrt{3}\ \text{cm}$$ intersecting at point E. A circle is drawn passing through three points A, E and B with its center at O. Straight-line EF is drawn parallel to AB such that the point F lies on the side BC. Also there lies a point P somewhere inside the rhombus then, 

This is a multiple-correct question, and all four options (a), (b), (c) and (d) are correct. The diagonals bisect at right angles, so $$AE = 5$$, $$BE = 5\sqrt{3}$$ and $$AB = BC = 10$$; as $$\angle AEB = 90^\circ$$, AB is a diameter, O is its midpoint and $$OE = 5$$, giving $$OE \colon BC = 1 \colon 2$$. Since E is the midpoint of AC and $$EF \parallel AB$$, F is the midpoint of BC, so $$FB = 5$$ cm. The rhombus has area $$50\sqrt{3}$$, triangle ABC has $$25\sqrt{3}$$, triangle EFC has $$\frac{1}{4}$$ of that, i.e. $$6.25\sqrt{3}$$, and trapezium AEFB has $$18.75\sqrt{3}$$. The probabilities are $$\frac{18.75}{50} = 0.375$$ and $$\frac{6.25}{50} = \frac{1}{8}$$.