I have been using Kaspersky Internet Security for a few years now and it kept my computer very clean. My daughter sometimes uses friends laptops with her thumb drive and it's always getting infected. Same with my wife and the college lab computers. I hated paying $50 every year and I switched to MSE a month ago. So far, it's kept my personal PC clean when they use it to print or edit their work. I double check with Dr. Web's free utility (link
- if the site is in some other language click English at top right corner).
MSE is not full proof and I've tested it with an old computer on some sites with malware in the advertisements as well as a few known black hat sites. It failed about 10% of the time, but usually only against the most vile stuff. If you don't surf porn sites, warez or use sharing programs like Limewire then you should be OK.
Some tips would be to make sure MSE is updated weekly (there's Optional updates every few days), use Firefox + AdBlock Plus and subscribe to "EasyList" and "EasyPrivacy". Those will kill most ad malware that tries to toss tracking cookies on your computer. Also, disable "Third Party Cookies" in Firefox. If that isn't enough, install NoScript. That way, you can add sites that you know are clean to the white list and let it ban scripts in sites you don't trust.
The one thing that NOD, Kaspersky, Nortons and McAfee's do is add a function called "Sandboxing". Basically, it allows you to run programs (manually or automatically) in their own protected memory space so if something does go wrong it will be contained. If this is for really non-techie people or teenagers, I would stay with a suite. Kids these days are always looking for songs, free movies, using limewire, porn and running all sorts of programs their friends send them. Not to mention, all the top viruses of the past traveled via open ports, not scripts/programs. For the more technical, MSE, Avast! or other free programs will probably be enough. Just whatever you do, do NOT use CA. They're one of the worst out there and rank even below the free solutions.