Posted by anonymous on Friday, October 19, 2007 at 9:10pm.
If 90% of the households in a certain region have an answering machines and 50% have both answering machines and call waiting, what is the probability that a household chosen at random and found to have an answering machine also has call waiting?
This is what I think...(Am I correct?)
P(answering and call waiting/answering machine)=(.50/.90)=.56 or 56%

math  bobpursley, Friday, October 19, 2007 at 10:02pm
DIdn't the probem state that 50 percent have both? Am I missing something here?
I don't read a conditional probabality on the fifty percent.

mathwhy not?  anonymous, Friday, October 19, 2007 at 10:29pm
have an "answering machine" also has call waiting?

math  bobpursley, Saturday, October 20, 2007 at 5:10am
OK,assumeing the problem statement is conditional, then you are correct. 56 percent of those who have answering machine have call waiting also.
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