Question

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10. At a certain pizza parlor , 56 % of the customers order a pizza containing onions , 55 % of the customers order a pizza containing sausage , and 69% order a pizza containing onions or sausage (or both ). Find the probability that a customer chosen at random will order a pizza containing both onions and sausage. Write your answer as a decimal (not as a percentage ). (If necessary , consult a list of formulas .)

15.

A sample of 279 adults is selected . The adults are classified according to voter registration status (" registered " or " not registered " ). They are also classified according to preferred source of current events information (" television " , " newspapers ", " radio ", or " internet sites " ). The results are given in the contingency table below . Internet sites Television Newspapers 30 37 42 15 Registered Not registered 35 47 42 What is the relative frequency of adults in the sample who prefer to obtain their information through newspapers ? Round your answer to two decimal places

Answer 1

Answer 10:

Data given:

At a certain pizza parlor - 56 % of the customers order a pizza containing onions, 55 % of the customers order a pizza containing sausage, and 69% order a pizza containing onions or sausage (or both).

Probability that a customer chosen at random will order a pizza containing both onions and sausage, = ?

Probability that a customer chosen at random will order a pizza containing onions,

Probability that a customer chosen at random will order a pizza containing sausage,

Probability that a customer chosen at random will order a pizza containing onions or sausage (or both),

As we know, for any two events A and B - the probability of the union of two events is given as -

Using the above information, we can find the required probability as -

Thus, the probability that a customer chosen at random will order a pizza containing both onions and sausage is 0.42 .