Calculating Probabilities for Dorm Room Appliances: TV, Fridge, or Neither?

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The discussion revolves around calculating probabilities related to dorm room appliances, specifically TVs and fridges. The user initially struggles with determining the probability of having a TV but no fridge, a TV or fridge but not both, and neither appliance. They receive feedback emphasizing the importance of considering the overlap of rooms with both appliances and the utility of using a Venn diagram for visualization. The conversation highlights the need to correctly apply probability formulas and the importance of posting in the appropriate forum section for homework-related questions. Overall, the user gains clarity on their calculations and the approach to solving the problem.
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So I'm having some trouble with a particular question in my stats textbook. It's an even number and the answers in the back cover only the odd questions so i can't check my answers. Here's the question:
Dorm rooms on a university campus reveal that 38% have a fridge, 52% a TV and 21 % have both a fridge and a TV. Find the probability that a dorm room has:
a)a TV but no fridge
b)A tv or fridge BOT NOT BOTH??
c)neither a tv nor a fridge

Heres what I did:
a)Muliplied the probability of having a tv by 1 minus the probability of having a fridge
.52 x (1-.32)= .35
b)Heres where I got confused. It says but not both? I've never seen that before. I just used the standard formula for p(aorb)=p(a)-p(b)-p(a&b) which is .52+.38-.21=.69 Is that right?
c)1 minus the probability of having a fridge or tv (the answer i got for b) so
1-.69= .31

Any feedback of whether I am doing this right would be greatly appreciated.
 
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Why can't i get any assistance with this question? Is it that stupid? Is it because I am a "noob" Seriously any feedback positive or negative would be GREATLY appreciated.
I realize now that i did b wrong the p(aorb) with the given formula includes the probability of p(a) and p(b) occurring simultaneously. Should I calculate with p(a&notb) x p(b&nota) for p(aorb) when they both cannot occur??
 
You're new here, so you probably didn't realize that homework problems should be posted in the Homework and Coursework section, not in the Math section.

Yes, you need to take into account the rooms that have both a fridge and a TV (21%). I drew a Venn diagram with two overlapping circles to figure out the probabilities. In the Venn diagram there are four distinct regions - rooms with a refrigerator (only), rooms with a TV (only), rooms with both a refrigerator and a TV, and rooms with neither.
 
Thanks mark. Ill be sure to post problems in the homework section from now on. You're right it helps to visualize the problem as a venn diagram instead of plugging numbers into formulas or trying to make new ones:p
 
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