Problem
The phrase “what distribution is the coat hangers problem probability” typically refers to a situation where a box contains a fixed number of coat hangers, some bent and some normal, and a sample is drawn without replacement.
Concrete version (fully specified): A box contains \(N=30\) coat hangers, of which \(K=8\) are bent. A student randomly selects \(n=5\) hangers without replacement. Let \(X\) be the number of bent hangers selected.
Tasks: (1) Identify the distribution of \(X\). (2) Write the probability mass function. (3) Compute \(P(X=2)\).
Step 1: Recognize the sampling structure
- There is a finite population of size \(N\).
- Each item is one of two types (bent = “success”, normal = “failure”).
- A sample of size \(n\) is drawn without replacement, so draws are not independent.
This structure matches the hypergeometric distribution.
Step 2: State the hypergeometric model
If \(X\) counts the number of “successes” (bent hangers) in the sample, then:
\[ X \sim \operatorname{Hypergeometric}(N, K, n) \]
| Symbol | Meaning | Value here |
|---|---|---|
| \(N\) | Total number of coat hangers in the box | 30 |
| \(K\) | Number of bent (success) hangers in the box | 8 |
| \(n\) | Number of hangers selected | 5 |
| \(X\) | Number of bent hangers in the sample | \(0,1,2,3,4,5\) |
Step 3: Probability mass function
The hypergeometric probability mass function is:
\[ P(X=x)=\frac{\binom{K}{x}\binom{N-K}{n-x}}{\binom{N}{n}} \quad \text{for } \max\!\bigl(0,\,n-(N-K)\bigr)\le x \le \min(n,K) \]
In this coat hangers problem, the support is \(x=0,1,2,3,4,5\).
Step 4: Compute a sample probability \(P(X=2)\)
“Exactly 2 bent hangers” means choosing 2 from the 8 bent and 3 from the 22 normal:
\[ P(X=2)=\frac{\binom{8}{2}\binom{22}{3}}{\binom{30}{5}} \]
Evaluate the combinations:
\[ \binom{8}{2}=28,\quad \binom{22}{3}=1540,\quad \binom{30}{5}=142506 \]
Substitute:
\[ P(X=2)=\frac{28\cdot 1540}{142506} =\frac{43120}{142506} =\frac{3080}{10179} \approx 0.3026 \]
Visualization: population split and the sampled count
Final conclusion
For a coat hangers problem probability question where hangers are selected without replacement from a box with a fixed number of bent hangers, the correct distribution is the hypergeometric distribution, and probabilities follow \[ P(X=x)=\frac{\binom{K}{x}\binom{N-K}{n-x}}{\binom{N}{n}}. \]