Solution: This is a binomial probability problem with $ n = 4 $ trials (rolls), success probability $ p = \frac16 $ (rolling a 4), and we want exactly $ k = 2 $ successes. - ToelettAPP
Solving a Binomial Probability Problem: Exactly 2 Successes in 4 Rolls with p = 1/6
Solving a Binomial Probability Problem: Exactly 2 Successes in 4 Rolls with p = 1/6
If youâÂÂve ever tossed a die and asked, âÂÂWhatâÂÂs the chance of rolling a 4 exactly twice in only 4 rolls?â â youâÂÂre dealing with a classic binomial probability problem. In this article, we break down how to solve this using the binomial distribution formula, with $ n = 4 $, success probability $ p = rac{1}{6} $, and exactly $ k = 2 $ successes.
Understanding the Context
What is a Binomial Probability Problem?
A binomial probability problem describes experiments consisting of $ n $ independent trials, each with two possible outcomes: success or failure. The probability of success $ p $ remains constant across trials, and we want to compute the probability of exactly $ k $ successes.
The binomial probability formula is:
$$
P(X = k) = inom{n}{k} p^k (1 - p)^{n - k}
$$
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Key Insights
Where:
- $ inom{n}{k} $: the binomial coefficient, representing the number of ways to choose $ k $ successes from $ n $ trials
- $ p $: probability of success on a single trial
- $ 1 - p $: probability of failure
Problem Setup
Given:
- Number of trials $ n = 4 $
- Success probability $ p = rac{1}{6} $ (rolling a 4 with a fair die)
- Number of desired successes $ k = 2 $
We want to find:
The probability of rolling exactly two 4s in four rolls of a fair die.
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Step-by-Step Solution
Step 1: Identify parameters
- $ n = 4 $ (four rolls)
- $ p = rac{1}{6} $ (rolling a 4)
- $ k = 2 $ (exactly two 4s)
- Thus, $ 1 - p = rac{5}{6} $ (probability of not rolling a 4)
Step 2: Compute the binomial coefficient
The number of ways to choose 2 successes (rolling a 4) from 4 trials is:
$$
inom{4}{2} = rac{4!}{2!(4-2)!} = rac{24}{2 \cdot 2} = 6
$$
Step 3: Apply the binomial formula
$$
P(X = 2) = inom{4}{2} \left( rac{1}{6}
ight)^2 \left( rac{5}{6}
ight)^{4 - 2}
$$
$$
P(X = 2) = 6 \cdot \left( rac{1}{6}
ight)^2 \cdot \left( rac{5}{6}
ight)^2
$$
$$
P(X = 2) = 6 \cdot rac{1}{36} \cdot rac{25}{36} = rac{6 \cdot 25}{36 \cdot 36} = rac{150}{1296}
$$
