Suppose that a box of 86 circuits is sent to a manufacturing plant. Of the 86 circuits shipped, 5 are defective. The plant manager receiving the circuits randomly selects 2, without replacement, and tests them. If both circuits work, the manager will accept the shipment. Otherwise the shipment is rejected. What is the probability that the shipment is accepted

the probability that the shipment is accepted is 0.8865

Step-by-step explanation:

Given the data in the question;

N = 86,  n/d = 5 and n = 2

now, without replacement

the probability that the shipment is accepted will be;

probability that the shipment is accepted = probability that non is defectives

so p( non is defective ) = ( (86-5)/86) × ((86-5-1)/(86-1))            

p( non is defective ) =  ( 81 / 86) × (80/85)

p( non is defective ) =  0.8865

Therefore, the probability that the shipment is accepted is 0.8865

answer: 1, 3 and 5 are correct

step-by-step explanation: