Final answer:
The probability that Carl's wife chooses sausage and onion is 2/21, or approximately 0.0952.
Step-by-step explanation:
To find the probability that Carl's wife chooses sausage and onion, we first need to determine the total number of possible combinations of 2 toppings. Since there are 7 toppings to choose from, we can calculate this using the formula for combinations: C(7, 2) = 7! / (2! * (7-2)!) = 21.
Next, we need to determine the number of combinations that include sausage and onion. Since we are choosing 2 toppings, there are 2 combinations that include both sausage and onion: (sausage, onion) and (onion, sausage).
Therefore, the probability that Carl's wife chooses sausage and onion is 2/21, or approximately 0.0952 (rounded to four decimal places).
Complete Question:
A pizza restaurant is offering a special price on pizzas with 2 toppings. The toppings they offer are: pepperoni, mushroom, sausage, green pepper, pineapple, ham, and onion. Suppose that Carl's favorite is sausage and mushroom, but his wife can't remember that, and she is going to randomly choose 2 different toppings. What is the probability that Carl's wife chooses sausage and onion?