Question: If I roll a fair six-sided die four times, what is the probability that I roll the number 4 exactly twice? - Simpleprint
Probability of Rolling Exactly Two 4s When Rolling a Die Four Times
Probability of Rolling Exactly Two 4s When Rolling a Die Four Times
Rolling a fair six-sided die four times is a classic probability scenario that many people encounter, whether in games, education, or just casual curiosity. A common question arises: If I roll a fair six-sided die four times, what is the probability that I roll the number 4 exactly twice? Understanding this probability involves applying the principles of binomial probability, making it a great example to explore how random events and combinations work.
Understanding the Binomial Probability Framework
Understanding the Context
This problem fits perfectly within the binomial distribution framework. The binomial distribution applies when:
- There are a fixed number of independent trials (here, 4 die rolls).
- Each trial has only two outcomes: รยขรยรยsuccessรยขรยรย (rolling a 4) or รยขรยรยfailureรยขรยรย (rolling anything other than 4).
- The probability of success remains constant per trial (for a fair die, P(4) = 1/6).
- Trials are independent.
In this context:
- Success = rolling a 4 (probability ( p = rac{1}{6} ))
- Failure = rolling not a 4 (probability ( q = 1 - p = rac{5}{6} ))
- Number of trials ( n = 4 )
- Desired number of successes ( k = 2 )
Step-by-Step Calculation of the Probability
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Key Insights
1. Calculate the number of favorable outcomes
We need the number of ways to roll exactly two 4s in four rolls. This is a combination problem:
[
inom{4}{2} = rac{4!}{2!(4-2)!} = rac{24}{2 \cdot 2} = 6
]
There are 6 unique sequences (e.g., 4,4,n,n in all combinations) where exactly two rolls show a 4.
2. Calculate the probability for one such sequence
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For any specific sequence with exactly two 4s and two non-4s (e.g., 4, 4, 2, 5), the probability is:
[
P = \left( rac{1}{6}
ight)^2 \ imes \left( rac{5}{6}
ight)^2 = rac{1}{36} \ imes rac{25}{36} = rac{25}{1296}
]
3. Multiply by the number of favorable sequences
Since the 6 arrangements are mutually exclusive, the total probability is:
[
P(\ ext{exactly 2 fours}) = inom{4}{2} \ imes \left( rac{1}{6}
ight)^2 \ imes \left( rac{5}{6}
ight)^2 = 6 \ imes rac{25}{1296} = rac{150}{1296}
]
4. Simplify the result
[
rac{150}{1296} = rac{25}{216} pprox 0.1157 \ ext{ or } 11.57%
]
Final Answer
The probability of rolling exactly two 4s when rolling a fair six-sided die four times is:
[
oxed{ rac{25}{216}} \quad \ ext{or approximately} \quad 11.57%
]
