Question 1184503: A teacher wants to select randomly a student from a group of 4 students. The names of the students are Anna, Maria, Alex, Ivan.
(a) Define the set of basic outcomes for this problem. What is the probability of each outcome?
(b) Consider the random events A = “the name of the selected student starts with A”, B = “the name of the selected student ends with a”. What outcomes do they consist of? What are their probabilities?
(c) Find the probabilities of the events A∪B, A∩B, A\B, (notA)∪(not B), (not A)∩(not B).
Consider the same 4 students as in the previous problem, but suppose that the teacher wants to select 2 students.
(a) Define the set of basic outcomes for this problem (you can abbreviate the names to write less). What is the probability of each outcome?
(b) Let A = “the name of at least one of the selected student starts with A”, B = “the name of at least one of the selected student endswith a”. Find the probabilities of these events.
(c) Find the probabilities of the events A∪B, A∩B.
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let's break down these probability problems step by step.
**Problem 1: Selecting 1 Student**
**(a) Basic Outcomes and Probabilities:**
* **Set of Basic Outcomes:** {Anna, Maria, Alex, Ivan}
* **Probability of each outcome:** Since the selection is random, each student has an equal chance of being chosen. Therefore, the probability of each outcome is 1/4 or 0.25.
**(b) Events A and B:**
* **Event A ("starts with A"):** {Anna, Alex} P(A) = 2/4 = 1/2 = 0.5
* **Event B ("ends with a"):** {Anna, Maria} P(B) = 2/4 = 1/2 = 0.5
**(c) Combined Probabilities:**
* **A ∪ B (A or B or both):** {Anna, Maria, Alex} P(A ∪ B) = 3/4 = 0.75 (This is because 3 out of the 4 students satisfy either condition A or B or both.)
* **A ∩ B (A and B):** {Anna} P(A ∩ B) = 1/4 = 0.25 (Only Anna satisfies both conditions.)
* **A \ B (A but not B):** {Alex} P(A \ B) = 1/4 = 0.25 (Only Alex starts with A but doesn't end with a.)
* **(not A) ∪ (not B) (Not A or Not B):** {Ivan, Maria, Alex} P((not A) ∪ (not B)) = 3/4 = 0.75. (This is the same as not(A and B), by De Morgan's Law)
* **(not A) ∩ (not B) (Not A and Not B):** {Ivan} P((not A) ∩ (not B)) = 1/4 = 0.25 (Only Ivan satisfies neither condition.)
**Problem 2: Selecting 2 Students**
**(a) Basic Outcomes and Probabilities:**
We'll use abbreviations: A = Anna, M = Maria, X = Alex, I = Ivan. Since we're selecting 2 students, the order doesn't matter (combinations, not permutations). The possible pairs are:
* {A, M}
* {A, X}
* {A, I}
* {M, X}
* {M, I}
* {X, I}
There are 6 possible outcomes. Since the selection is random, each outcome has a probability of 1/6.
**(b) Events A and B:**
* **Event A ("at least one starts with A"):** {A, M}, {A, X}, {A, I} P(A) = 3/6 = 1/2 = 0.5
* **Event B ("at least one ends with a"):** {A, M}, {M, X} P(B) = 2/6 = 1/3 ≈ 0.333
**(c) Combined Probabilities:**
* **A ∪ B (at least one starts with A *or* at least one ends with a):** {A, M}, {A, X}, {A, I}, {M, X}. P(A ∪ B) = 4/6 = 2/3 ≈ 0.667
* **A ∩ B (at least one starts with A *and* at least one ends with a):** {A, M} P(A ∩ B) = 1/6 ≈ 0.167
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