Probability regarding 3 random toppings on a pizza At Luigi’s House of Random Pizza, you can order only one thing: A medium pizza with three
random topping layers chosen by Luigi himself. Luigi has the following toppings: Pepperoni,
Pineapple, Mushroom, Anchovie, Onion, and Jalepeno Peppers. For example, you might get:
Pepperoni-Onion-Mushroom, or maybe Mushroom-Pepperoni-Onion (the order of the layers mat-
ters to Luigi) or if you are especially lucky then Anchovie-Anchovie-Anchovie. I am not sure if my thought process is correct for both of these.

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*What is the probability that you get a pizza with either two or three distinct toppings? (For
example, Mushroom-Onion-Mushroom or Pepper-Onion-Mushroom.)

What I tried for this is that the probability for more than 1 topping would be equal to: $1 - P(1~\text{topping}) = 1 - \dfrac{6}{6^3} = 0.97$.


*What is the probability that Luigi gives you a pizza with exactly two distinct toppings
(occurring in any order)?

I am unsure of my logic for this but I believe that the equation would be: $P(2~\text{toppings}) = 1 - P(1~\text{topping}) - P(3~\text{toppings}) = 0.97 - P(3~\text{toppings})$. What would I do to find the probability for $P(3~\text{toppings})$ in order to use the equation to solve for $P(2~\text{toppings})$?
 A: Your first answer is correct. Let the first toppings be any one of the six. The probability that the remaining two toppings are the same as the first is $~ \displaystyle \frac 16 \cdot \frac 16 = \frac 1{36}$
So the probability that there are two or three distinct toppings is,
$ \displaystyle  1 - \frac{1}{36} = \frac{35}{36}$, same as you obtained.
For the second question, note that there are two cases - i) if the second topping is same as the first then the third has to be from the remaining $5$ toppings ii) if the second topping is different than the first then the third topping is either from the first or the second. So the probability is,
$ \displaystyle \frac 16 \cdot \frac 56 + \frac 56 \cdot \frac 26 = \frac{5}{12}$
You could have also subtracted the probability of three distinct toppings from the answer to the first question. The probability of three distinct topping is,
$ \displaystyle \frac 56 \cdot \frac 46 = \frac 59~$ and
$ \displaystyle \frac {35}{36} - \frac 59 = \frac{5}{12}$
