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We throw a die 10 times. What's the probability of getting at least one 6?

Using the complentary that's $1-Pr$(Getting no $6$'s)

$1- ({\frac{5}{6}})^{10} = 0.838494$

There's a note that says that to solve the problem directly would require a complex use of the additivity property.

a) How do you solve the problem directly?

I think that to solve it directly, though not sure, is to find the $P(A_1 \cup A_2 \cup A_3 \cup A_4 \cup A_5 \cup A_6)$ in ten die throws, where $A_i$ is the event of getting $i$ $6$'s

As the events $A_i$ are not disjoint the exclusion-inclusion principle is needed.

My try:

$$\binom{10}{1} \frac{1}{6} - \binom{10}{2} \frac{1}{6^2} + \binom{10}{3} \frac{1}{6^3} - \binom{10}{4} \frac{1}{6^4}+ \binom{10}{5} \frac{1}{6^5} - \binom{10}{6} \frac{1}{6^6} =0.838091$$

It is a slightly different result that the one obtained using the complementary. What am I not doing right?

b)If the events aren't disjoint, can you talk about the "additivity property"? As it is defined for disjoint events.

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It would be the following:

$$\sum_{n=1}^{10} \dbinom{10}{n}\left(\dfrac{1}{6}\right)^n\left(\dfrac{5}{6}\right)^{10-n}$$

Here the events are disjoint. You have exactly 1 roll of a 6, exactly two rolls of a six, ..., exactly ten rolls of a six.

Your attempt works as well (although you stopped at $i=6$). Had you continued using Inclusion/Exclusion up to $i=10$, it would have given the same answer:

$$\sum_{i=1}^{10}(-1)^{i+1}\dbinom{10}{i}\left(\dfrac{1}{6}\right)^i$$

Here you are counting if you choose a die, the probability that one die is a six, minus if you choose two dice, both of them are a six, plus if you choose three dice, all three are a six, etc.

b) Yes.

$$P(A\cup B) = P(A)+P(B)-P(A\cap B)$$

This is the property that yields the Inclusion/Exclusion principle in the first place.

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You got a different answer because you only computed the first six terms of the inclusion-exclusion method. You have ten dice, each of which could produce a six, so you need ten terms.

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  • $\begingroup$ You use Inclusion-Exclusion when you have events that are not mutually exclusive. You can see that "roll six on the first die" and "roll six on the second die" are not mutually exclusive because it is possible that you roll sixes on both dice. I don't know what you mean by "inclusive." $\endgroup$ – David K May 3 at 21:27

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