A consignment of 20 picture tubes contains 5 defectives. Two tubes are selected one after the other at random. The probability that both are defective assuming that the first tube is not replaced before drawing
the second, would be:	Pick one:
A) 1/16
B) 1/19
C) 1/4
D) 1/3
E) none of these
Probability of drawing a defective tube in first draw is 5/20   Probability of drawing a defective tube in second draw (without replacing first tube) is 4/19.   Therefore probability of getting both defective is = (5/20)*(4/19) = 1/19   ANSWER:1/19
<result>B</result>
A consignment of 20 picture tubes contains 5 defectives. Two tubes are selected one after the other at random. The probability that both are defective assuming that the first tube is not replaced before drawing
the second, would be: