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%
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.
have an "answering machine" also has call waiting?
OK,assumeing the problem statement is conditional, then you are correct. 56 percent of those who have answering machine have call waiting also.
To calculate the probability that a household chosen at random and found to have an answering machine also has call waiting, you can use conditional probability.
Conditional probability is the probability of an event occurring given that another event has already occurred. In this case, the event of interest is the probability of a household having call waiting, given that they already have an answering machine.
To calculate this, you would divide the number of households that have both call waiting and answering machines by the number of households that have answering machines.
In this case, you've been given that 90% of households have answering machines and 50% of households have both answering machines and call waiting.
So, to calculate the probability, you would divide 50% (the number of households with both call waiting and answering machines) by 90% (the number of households with answering machines).
This can be expressed as:
P(call waiting given answering machine) = P(call waiting and answering machine) / P(answering machine)
P(call waiting given answering machine) = 0.50 / 0.90
Simplifying this expression, you get:
P(call waiting given answering machine) ≈ 0.5556 or approximately 55.56%
Therefore, the correct answer is approximately 55.56% or 0.5556.