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My stats test is tomorrow, and my professor emailed me a question to consider.

Ben gets a three-scoop ice cream cone. There's a .4 probability that the middle scoop is chocolate, .2 prob that the top scoop is chocolate, and .6 prob that the bottom scoop is chocolate. The different scoops are mutually independent.

What's the probability that Ben gets a cone with exactly one scoop of chocolate? (How about exactly two scoops?)

For one scoop, I initially multiplied the probs to get .048, but that seems wrong. I think I need to create a union. $(T\cap NB \cap NM) \cup (NT \cap NB \cap M) \cup (NT \cap B \cap NM).$

T=Top, B = Bottom, M = Middle, NT = Not Top, etc.

I'm not sure how to get the values, assuming this is correct. Do I need to create a Venn Diagram with three circles? How? A intersects B in 2 places, etc.

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  • $\begingroup$ By the way, $0.048$ is the probability of getting exactly three scoops of chocolate. $\endgroup$ – Graham Kemp Oct 19 '16 at 2:34
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You have a union of disjoint events; the probability of that is the sum of probabilities of the events.

Each of those events is an intersection of (mutually) independent events; the probability of such is the product of the probabilities of the independent events.

$$\begin{align} &\mathsf P((A\cap B^c\cap C^c)\cup(A^c\cap B\cap C^c)\cup(A^c\cap B^c\cap C))\\ =~& \mathsf P(A)\mathsf P(B^c)\mathsf P(C^c)+\mathsf P(A^c)\mathsf P(B)\mathsf P(C^c)+\mathsf P(A^c)\mathsf P(B^c)\mathsf P(C)\end{align}$$

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  • $\begingroup$ Other than recognizing that .6+.4+.2 must be wrong (because it totals to 1.2), how do you know to add this union, but with three scoops you multiply? $\endgroup$ – Ham Sandwich Oct 19 '16 at 2:39
  • $\begingroup$ @T-1000'sSon You always add with a union and multiply with an intersection. If the union is of disjoint events then just add, otherwise use the Principle of Inclusion and Exclusion. If the intersection is of mutually independent events, just multiply, otherwise use conditional probability. $\endgroup$ – Graham Kemp Oct 19 '16 at 3:27

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