Two assembly lines I and II have the same rate of defectives in their production of voltage regulators. Six regulators are sampled from each line and tested. Among the total of twelve tested regulators, four are defective. Find the probability that exactly two of the defective regulators came from line I. (Round your answer to four decimal places.)

0.4546

Step-by-step explanation:

nCr = n!/(n-r)!r!

Number of ways of selecting the four defective voltage regulators from 12 =  12C4 = 12!/(12-4)!4! = 12!/8!4! = (12 *11*10*9)/(4*3*2*1)

12C4 =  495 ways

Number of ways of selecting 2 defectives from line 1 = 6C2 * 6C2

6C2 = 6!/(6-2)!2! = 6!/4!2! = (6*5)/(2*1) = 15

6C2 * 6C2 = 15*15 = 225 ways

Probability = Number of possible outcomes/ Number of total outcomes

Probability that exactly 2 of the defective regulators came from line 1 = 225/40.95 = 0.4546

on monday rodney ate 1/4th of the sandwich = 3/12

on tuesday he ate 1/3rd of the sandwich = 4/12

therefore in total he has eaten = 3/12 + 4/12 = 7/12

thus we can say that he has eaten 7/12 of the sandwich on monday and tuesday alltogether.

happy to

pls mark as brainliest.