Answer:
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%
