The third proportional to 0.36 and 0.48 is
Due to some force F1 a body oscillates with period 4/5 sec and due to other force F2 oscillates with period 3/5 sec. If both forces act simultaneously, the new period will be
A and B are events such that P(A) = 0.42, P(B) = 0.48 and P(A and B) = 0.16. Determine (i) P (not A ) (ii) P (not B) and (iii) P (A or B)