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A coin has two sides: heads and tails. A die has six sides, numbered 1 through 6. If you flip the coin and roll the die, what is the probability that you flip "heads" and roll a number greater than 5?

A. [tex]\(\frac{1}{12}\)[/tex]
B. [tex]\(\frac{5}{6}\)[/tex]
C. [tex]\(\frac{1}{6}\)[/tex]
D. [tex]\(\frac{1}{4}\)[/tex]

Sagot :

To solve the problem of finding the probability that you flip "heads" on a coin and roll a number greater than 5 on a die, we need to break the problem down into steps.

1. Determine the probability of flipping "heads" on the coin:
- A standard coin has two sides, "heads" and "tails". Therefore, the probability of flipping "heads" is:
[tex]\[ P(\text{heads}) = \frac{1}{2} \][/tex]

2. Determine the probability of rolling a number greater than 5 on the die:
- A standard die has six sides, numbered 1 through 6.
- The numbers greater than 5 are just 6.
- There is only 1 favorable outcome (rolling a 6) out of 6 possible outcomes. Therefore, the probability of rolling a number greater than 5 is:
[tex]\[ P(\text{number} > 5) = \frac{1}{6} \][/tex]

3. Calculate the combined probability of flipping "heads" and rolling a number greater than 5:
- Since the coin flip and the die roll are independent events, the combined probability is the product of the individual probabilities:
[tex]\[ P(\text{heads and number} > 5) = P(\text{heads}) \times P(\text{number} > 5) = \frac{1}{2} \times \frac{1}{6} \][/tex]
- Simplify the multiplication:
[tex]\[ P(\text{heads and number} > 5) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12} \][/tex]

Therefore, the probability that you flip "heads" and roll a number greater than 5 is [tex]\(\frac{1}{12}\)[/tex].

The correct answer is:
[tex]\[ \boxed{\frac{1}{12}} \][/tex]
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