At a certain pizza parlor, 43% of the customers order a pizza containing onions, 33% of the customers order a pizza containing sausage, and 67% 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.



9%

Step-by-step explanation:

The percentage that a customer request for a pizza that contains onion = 43%

The percentage that a customer request for a pizza that contains sausage = 33%

The percentage that a customer request for a pizza that contains onion and/or sausage = 67%

We can represent this using probability notations

P(Onions) = 43% = 0.43

P(Sausage) = 33% = 0.33

P(Onions and/or Sausage) = 67% = 0.67

P(Onions and/or Sausage) can be translated to customers that request for pizza containing "Onions only" , "Sausage only", "Onions and Sausage"

Mathematical, we represent the above statement:

P(Onions) + P(Sausages) - P(Onions and Sausage)

So, we have

P(Onions and/or Sausage) = P(Onions) + P(Sausages) - P(Onions and Sausage)

0.67 = 0.43 + 0.33 - P(Onions and Sausage)

0.67= 0.76 - P(Onions and Sausage)

P(Onions and Sausage) = 0.76 - 0.67

P(Onions and Sausage) = 0.9

So, the probability of a customer making an order of pizza that contains onions and sausage is 0.9 or 9%

answer: 40.95

step-by-step explanation:

she gotta workk that a 2

step-by-step explanation: