SOLUTION: Please help me solve this question.
A drawer contains 2 black socks and 2 green socks. The socks are removed randomly, one at a time, from the drawer until 2 socks of the same col
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Question 1171470: Please help me solve this question.
A drawer contains 2 black socks and 2 green socks. The socks are removed randomly, one at a time, from the drawer until 2 socks of the same colour are removed. The random variable X is the number of socks removed from the drawer.
a. Tabulate the probability distribution P(X = x). (
b. Hence, calculate the mean and variance of X.
Answer by CPhill(2189) (Show Source): You can put this solution on YOUR website!
Let's break down this problem step by step.
**a) Tabulating the Probability Distribution P(X = x)**
* **Possible Values of X:**
* X = 2: Two socks are removed, and they are the same color.
* X = 3: Three socks are removed, and the first two are different colors.
* **Calculating Probabilities:**
* **P(X = 2):**
* There are 4 socks in total.
* The first sock can be any color (black or green).
* If the first sock is black, there's 1 black sock left out of 3. So the probability of drawing another black sock is 1/3.
* If the first sock is green, there's 1 green sock left out of 3. So the probability of drawing another green sock is 1/3.
* Probability of drawing two black socks: (2/4) * (1/3) = 2/12 = 1/6
* Probability of drawing two green socks: (2/4) * (1/3) = 2/12 = 1/6
* Therefore, P(X = 2) = 1/6 + 1/6 = 2/6 = 1/3
* **P(X = 3):**
* This means the first two socks are different colors, and the third sock matches one of the first two.
* Probability of drawing a black sock then a green sock: (2/4) * (2/3) = 4/12 = 1/3
* Probability of drawing a green sock then a black sock: (2/4) * (2/3) = 4/12 = 1/3
* Thus probability of first two being different is 1/3+1/3 = 2/3.
* Regardless of the first two socks, the third sock must match one of them. There are two socks left that would match, out of the remaining two socks. So the probability of the third sock matching is 2/2 = 1.
* Therefore, P(X = 3) = 2/3 * 1 = 2/3.
* **Probability Distribution Table:**
| x | P(X = x) |
|---|----------|
| 2 | 1/3 |
| 3 | 2/3 |
**b) Calculating the Mean and Variance of X**
* **Mean (E[X]):**
* E[X] = Σ [x * P(X = x)]
* E[X] = (2 * 1/3) + (3 * 2/3)
* E[X] = 2/3 + 6/3 = 8/3 ≈ 2.67
* **Variance (Var[X]):**
* Var[X] = E[X²] - (E[X])²
* E[X²] = Σ [x² * P(X = x)]
* E[X²] = (2² * 1/3) + (3² * 2/3)
* E[X²] = (4 * 1/3) + (9 * 2/3)
* E[X²] = 4/3 + 18/3 = 22/3
* Var[X] = 22/3 - (8/3)²
* Var[X] = 22/3 - 64/9
* Var[X] = (66/9) - (64/9) = 2/9 ≈ 0.22
**Summary:**
* **Mean (E[X]):** 8/3 ≈ 2.67
* **Variance (Var[X]):** 2/9 ≈ 0.22
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