Probability Theory Problem

0.5

I know nothing of sex refrigerators but I'm always up for the challenge.

Given that two tests have been carried out you require the probability than the remaining defective unit is in the third or fourth sex refrigerator.

The probability of achieving what some might describe as a defective unit in either of the first two of the remaining 4 sex refrigerators is :

1 chance in 4 the 3rd sex refrigerator will malfunction plus the chance that the 3rd will not malfunction (3 chances in 4) times the chance that the 4th will ( 1 chance in 3).

In total 1 chance in 4 of a defective 3rd unit plus 1 chance in 4 of a defective 4th unit equals o.5.

Please advice names and where I can visit the sex refrigerators.