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.

- math--why 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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