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A fair die is thrown three times:

  • What is the probability of getting: three sixes?
  • What is the probability of getting: six, one, six?

My solution:

Probability of getting three sixes:

$$\Pr(\text{1st dice six}) + \Pr(\text{2nd six}) + \Pr(\text{3rd six})$$ $$ = \frac{1}{6} \frac{5}{6} \frac{5}{6} + \frac{5}{6} \frac{1}{6} \frac{5}{6} + \frac{5}{6} \frac{5}{6} \frac{1}{6} = 3 \cdot \frac{25}{216} = \frac{75}{216} = \frac{25}{72}$$

Probability of getting six, one, six:

$$\Pr(\text{1st dice six}) + \Pr(\text{2nd one}) + \Pr(\text{3rd six})$$ $$ = \frac{1}{6} \frac{5}{6} \frac{5}{6} + \frac{5}{6} \frac{1}{6} \frac{5}{6} + \frac{5}{6} \frac{5}{6} \frac{1}{6} = 3 \cdot \frac{25}{216} = \frac{75}{216} = \frac{25}{72}$$

Am I on the right track?


Update:

The problem am trying to solve does not specify order. Can we assume that order matters?

How about the following question, would the result be different from the initial question?

A fair die is thrown three times:

  • What is the probability of getting: three sixes, where the first throw MUST be six?
  • What is the probability of getting: six, one, six, where the first throw MUST also be six?
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    $\begingroup$ If I understand the question correctly, the first probability would be $\left(\frac{1}{6}\right)^3$, that is, Pr(1st dice six and 2nd six and 3rd six). The probability of getting six, one, six (the order matters) would be the same. $\endgroup$ – Guest May 4 '15 at 7:33
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Am I on the right track?

No.

You've calculated: $$\mathsf P\left({\substack{(\text{first die is 6 and second and third are not 6})\text{ or }\\(\text{second die is 6 and first and third are not 6})\text{ or }\\(\text{third die is 6 and first and second are not 6})}}\right)$$

You should be looking for $$\mathsf P\Big(\text{first die is 6 and second die is 6 and third die is 6}\Big)$$


Likewise you've calculated: $$\mathsf P\left({\substack{(\text{first die is 6 and second and third are not 6})\text{ or }\\(\text{second die is 1 and first and third not are 1})\text{ or }\\(\text{third die is 6 and first and second are not 6})}}\right)$$

You should be looking for $$\mathsf P\Big(\text{first die is 6 and second die is 1 and third die is 6}\Big)$$


Hint:

  • Use the Product Rule for the probability of the conjunction of independent events. ("Conjunction" is "and", "intersection", et cetera.)

  • Use the Additive Rule for the probability of the disjunction of exclusive events (also called "disjoint events"). ("Disjunction" is "or", "union", et cetera)

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For your Update, consider this:, what is the probability that the first die is a six when it is given that the first die is a six?

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I am not sure what are you computing. But based on the statement of your problem. The total no. of outcomes are $6^3$.

For first case:

The only favorable outcome is $(6,6,6)$ thus the probability is $\frac{1}{216}$.

For second case:

The favorable outcomes are $(6,6,1), (6,1,6), (1,6,6)$ thus the probability is $\frac{3}{216}=\frac{1}{72}$.

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  • $\begingroup$ The way the question is formulated is unclear. Why say "six, one, six" if the order is not to matter? $\endgroup$ – Guest May 4 '15 at 7:39

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