A random pizza with exactly three toppings

Question

A pizza shop offers 8 toppings. A prankster orders a "random" pizza: for each of the 8 toppings, a fair coin decides independently whether it goes on. So every possible subset of toppings, including none and all eight, is equally likely.

What is the probability the pizza ends up with exactly 3 toppings?

Accepted Answer

Each topping is on or off, so the number of equally likely pizzas is 2^8 = 256. The pizzas with exactly 3 toppings correspond to choosing which 3 of the 8 toppings are on: \binom{8}{3} = 56. Therefore P(	ext{exactly 3 toppings}) = \frac{56}{256} = \frac{7}{32} \approx 0.219. Takeaway. Flipping an independent fair coin per item makes all 2^n subsets equally likely, so "exactly k chosen" is \binom{n}{k}/2^n, the symmetric binomial with p=	frac12. The counts \binom{n}{k} peak in the middle, so a moderate-size subset is most probable.

A random pizza with exactly three toppings

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