- #1
Notoriousb3
- 5
- 0
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.
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.