- The sum of all probabilities in a discrete distribution is equal to 1: \(P(X=0) + P(X=1) + P(X=2) = 1\).
- Substitute the given values: \(\frac{1}{2} + \frac{1}{3} + P(X=2) = 1\).
- Simplify the sum: \(\frac{3}{6} + \frac{2}{6} = \frac{5}{6}\).
- Calculate the final value: \(P(X=2) = 1 - \frac{5}{6} = \boxed{\frac{1}{6}}\).
Explanation
- Understanding the probability distribution property
The problem presents a probability distribution for a random variable \(X\), which represents the number of errors found in a machine. A fundamental property of any valid probability distribution for a discrete random variable is that the sum of the probabilities of all possible outcomes must equal \(1\).
The outcomes provided are:
- \(P(X = 0) = \frac{1}{2}\)
- \(P(X = 1) = \frac{1}{3}\)
- \(P(X = 2) = p\), where \(p\) is the value we need to determine.
Mathematically, this relationship is expressed as:
\[P(X=0) + P(X=1) + P(X=2) = 1\]
- Setting up and solving the equation
To find the missing probability \(p\), we substitute the given values into our equation:
\[\frac{1}{2} + \frac{1}{3} + p = 1\]
To solve this, we first find a common denominator for the fractions \(\frac{1}{2}\) and \(\frac{1}{3}\), which is \(6\):
\[\frac{3}{6} + \frac{2}{6} + p = 1\]
\[\frac{5}{6} + p = 1\]
- Calculating the final value
Finally, we isolate \(p\) by subtracting \(\frac{5}{6}\) from both sides of the equation:
\[p = 1 - \frac{5}{6}\]
\[p = \frac{6}{6} - \frac{5}{6}\]
\[p = \frac{1}{6}\]
Thus, the missing probability for \(X=2\) is \(\frac{1}{6}\).
Examples
In quality control at a textile factory, understanding the probability of machine errors allows managers to predict maintenance schedules. If we know the likelihood of zero or one error, we can deduce the likelihood of two or more errors, helping us allocate resources effectively. By ensuring the total probability sums to 1, we confirm the integrity of our statistical model, which is foundational for making data-driven decisions in operations.