$\begin {array} {1 1} (A)\;\large\frac{1}{2} & \quad (B)\;\large\frac{7}{15} \\ (C)\;\large\frac{2}{15} & \quad (D)\;\large\frac{1}{15} \end {array}$

PROBABILITY = no. of possible outcomes/ttotal number of outcomes

so here probabllity= 8c3/10c3=56/120=7/15