Question 1195546: Three different phone brands are selected, one at a time from an electronic store containing 6 Samsung phones, 6 I-phones and 6 Techno phones. let X denote the number of I-phones selected in 3 random selections from the store.
a. Express the probability distribution of X in tabular form.
b. What is the probability that at least one phone selected will be an I-phone?
c. what is the probability that none of three selected phones will be an I-phone?
d. what is the expected number of I-phone selected?
e. find variance of X and expected value of (4x-2)
Answer by ElectricPavlov(122)   (Show Source):
You can put this solution on YOUR website! Certainly, let's analyze the probability distribution of selecting I-phones.
**a. Probability Distribution of X in Tabular Form**
* **Possible Values of X:**
* X can take on values 0, 1, 2, or 3.
* **Total Phones:** 6 Samsung + 6 iPhones + 6 Techno = 18 phones
* **Calculate Probabilities:**
* **P(X = 0):** Probability of selecting no iPhones:
* (12/18) * (11/17) * (10/16) = 0.2710
* **P(X = 1):** Probability of selecting exactly one iPhone:
* (6/18) * (12/17) * (11/16) + (12/18) * (6/17) * (11/16) + (12/18) * (11/17) * (6/16) = 0.4662
* **P(X = 2):** Probability of selecting exactly two iPhones:
* (6/18) * (5/17) * (12/16) + (6/18) * (12/17) * (5/16) + (12/18) * (6/17) * (5/16) = 0.2268
* **P(X = 3):** Probability of selecting all three iPhones:
* (6/18) * (5/17) * (4/16) = 0.0357
* **Probability Distribution Table:**
| X (Number of iPhones) | P(X) |
|---|---|
| 0 | 0.2710 |
| 1 | 0.4662 |
| 2 | 0.2268 |
| 3 | 0.0357 |
**b. Probability of at least one iPhone**
* P(X ≥ 1) = P(X = 1) + P(X = 2) + P(X = 3)
* P(X ≥ 1) = 0.4662 + 0.2268 + 0.0357 = 0.7287
**c. Probability of no iPhones**
* P(X = 0) = 0.2710 (already calculated in part a)
**d. Expected Number of iPhones (E[X])**
* E[X] = Σ [X * P(X)]
* E[X] = (0 * 0.2710) + (1 * 0.4662) + (2 * 0.2268) + (3 * 0.0357)
* E[X] = 0 + 0.4662 + 0.4536 + 0.1071
* E[X] = 1
**e. Variance of X (Var[X])**
* Var[X] = E[X²] - (E[X])²
* E[X²] = Σ [X² * P(X)]
* E[X²] = (0² * 0.2710) + (1² * 0.4662) + (2² * 0.2268) + (3² * 0.0357)
* E[X²] = 0 + 0.4662 + 0.9072 + 0.3213
* E[X²] = 1.6947
* Var[X] = 1.6947 - (1)² = 0.6947
**Expected Value of (4X - 2)**
* E[4X - 2] = 4 * E[X] - 2
* E[4X - 2] = 4 * 1 - 2 = 2
**Summary:**
* Probability Distribution Table (see part a)
* P(X ≥ 1) = 0.7287
* P(X = 0) = 0.2710
* E[X] = 1
* Var[X] = 0.6947
* E[4X - 2] = 2
I hope this comprehensive explanation is helpful!
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