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  1. Brandon has 3 chances to roll a single die. He rolls 2 '6's, and a ‘5’ on his turn. Assuming the dice are fair, what is the probability of (i) ‘6’s on the first two rolls, and a non-six in the third roll, and (ii) getting exactly two ‘6’s?

On first glance, I calculated my answer to be i) 1/6 ^3 and ii) 1/6 ^2. Is this right? I think that each event (roll) is independent.

  1. In a game of Monopoly, 2 (fair) six-sided dice are rolled. Every face of each die is labelled 1 to 6. A double occurs if both dice land up on the same number. Instead of rolling both dice together, Brandon rolls one die at a time. The first die lands on a ‘6’. What is the probability that he rolls a double? If the first die lands on either ‘1’, ‘2’, ‘3’, ‘4’, ‘5’ or ‘6’, what is the probability that he rolls a double?

My answers to both the questions are 1/6, because I think that the first event where he rolls a specific number, would make up a P of 1. Thus we only need to calculate the second roll. Am I thinking in the right direction?

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    $\begingroup$ You are thinking correctly in both exercises, but in the first one there is an error. $(\frac{1}{6})^2$ is the probability of rolling twice a $6$ on the first two rolls. But what is a probability of not rolling a $6$ on the third roll? Furthermore in par ii) you only looked at the probability of rolling two sixes in two rolls. But the exercise asks about rolling exactly two sixes (in what order does not matter) in $3$ rolls. $\endgroup$ – Kore-N Feb 12 '17 at 13:08
  • $\begingroup$ I think that in number 1, both probabilities are equal to $1$. You are already $100\%$ sure he rolled `6's on the first two rolls and a non-6 on the third roll. $\endgroup$ – Joel Reyes Noche Feb 12 '17 at 13:28
  • $\begingroup$ @JoelReyesNoche Yeah i get what you mean :) But 1,1 is not in the options. Assuming that we ignore that sentence, what do you think is the answer? $\endgroup$ – Loki123 Feb 12 '17 at 13:31
  • $\begingroup$ You should have asked questions 1 and 2 in two separate posts. $\endgroup$ – Joel Reyes Noche Feb 12 '17 at 13:34
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You are correct that die rolls are assumed to describe events that are independent of each other. Recall that for independent events, the probability of event $E_1$ and event $E_2$ happening is the product of their probabilities. Also recall that for disjoint events, the probability of event $E_1$ or event $E_2$ happening is the sum of their probabilities.

For number 1, if we assume that the fair die is 6-sided and if we assume that die rolls are unknown (that is, if we ignore the statement that he has rolled a 6, 6, and a 5), then getting a 6 on the first roll (a $\frac16$ probability), and a 6 on the second roll (a $\frac16$ probability), and a non-6 on the third roll (a $\frac56$ probability) has a probability of $\frac16\cdot\frac16\cdot\frac56=\frac{5}{216}$.

Continuing number 1, getting exactly two 6's means either getting a 6, a 6, then a non-6, or getting a 6, a non-6, then a 6, or getting a non-6, a 6, then a 6. The probability of this is $\frac16\cdot\frac16\cdot\frac56+\frac16\cdot\frac56\cdot\frac16+\frac56\cdot\frac16\cdot\frac16=\frac{5}{216}+\frac{5}{216}+\frac{5}{216}=\frac{15}{216}=\frac{5}{72}$.

For number 2, your answers are correct. In the first instance, given that the first die roll is a 6, there is a $\frac16$ probability of getting a 6 in the second roll, that is, rolling a double. In the second instance, there are $36$ possible outcomes of rolling two dice. Of these, $6$ are doubles. The probability is thus $\frac{6}{36}=\frac16$.

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