Question 2
1 pts
Two fair dice are rolled. What is the probability that you will first roll a
number greater than 2, and then roll a 6?
1/4
3/22
1/9
5/64
Desk 1
EED
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Answer :

To determine the probability of two specific events happening in sequence when rolling two fair dice, we need to calculate the probability of each event individually and then combine them using the rules of probability.

### Step-by-Step Solution:

1. First Event: Rolling a number greater than 2 on a fair die

- A fair die has 6 faces, numbered from 1 to 6.
- The numbers greater than 2 are 3, 4, 5, and 6.
- So, there are 4 favorable outcomes out of a total of 6 outcomes.

Thus, the probability of rolling a number greater than 2 is:
[tex]\[ \text{Probability (number > 2)} = \frac{4}{6} = \frac{2}{3} \approx 0.6667 \][/tex]

2. Second Event: Rolling a 6 on a fair die

- A fair die also has 6 faces, and there is only one face showing the number 6.
- So, there is 1 favorable outcome out of a total of 6 outcomes.

Thus, the probability of rolling a 6 is:
[tex]\[ \text{Probability (rolling a 6)} = \frac{1}{6} \approx 0.1667 \][/tex]

3. Combined Probability: Both events happening in sequence

- For the combined probability of two independent events happening in sequence, we multiply the probabilities of the two individual events.

Thus, the combined probability is:
[tex]\[ \text{Combined Probability} = \left(\frac{2}{3}\right) \times \left(\frac{1}{6}\right) = \frac{2 \times 1}{3 \times 6} = \frac{2}{18} = \frac{1}{9} \approx 0.1111 \][/tex]

### Final Result:

The probability that you will first roll a number greater than 2 and then roll a 6 is [tex]\(\boxed{\frac{1}{9}}\)[/tex].

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