Show that if each shell has a 50% chance of being a dud, using the testing procedure, there is a 3/4 probability that the full lot if 399 or 398 shells will be accepted.

please explain fully.

The shell were manufactured in Lots, of four hundred, each Lot being subdivided into Sub-lots of one hundred. When a Lot was brought forward by the shell-maker for Proof, two shells were picked out at random from Sub-lot No. 1, to be tested at an armour plate of specified thickness, at a specified striking velocity and at a specified angle of impact. If the first shell succeeded in penetrating the plate whole, the full Lot of (now 399) shell passed into service. If this first shell failed the second was fired. If this was successful, the full lot of (now) 398 were accepted. If the second shell also failed, the Sub-lot was sentenced "Reproof' and the shell maker was given the option of withdrawing the whole lot from Proof or allowing the remaining 3 Sub-lots to proceed to Proof. Needless to say, they generally chose the latter. Proof then commenced with the next Sub-lot of 100, acceptance being governed by the same procedure as that for Sub-lot No. 1. The Ordnance Board's Professor of Statistics, having been given the results of Proof firings of the heavy shell then in the fleet, calculated that from 30 percent to 70 percent were probably dud shell, but the data from these Proof firings was insufficient to enable him to give a nearer approximation