Solving Elementary Statistics Problems: Simple Probability

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The discussion centers on a student struggling with basic probability concepts in an elementary statistics course. The student seeks help with two specific problems involving the probability of customers purchasing gas or electric dryers. For the first problem, the student deduces that the probability of at least two customers buying electric dryers is 0.572, based on the complement rule. In the second problem, the student grapples with understanding how to calculate the probability of at least one of each type being purchased, recognizing that the sum of probabilities for all possible outcomes must equal one. The conversation highlights the challenges of visualizing probability scenarios and applying fundamental equations to real-world situations.
exitwound
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I'm new to Statistics. Never had it in High School and now I'm in Elementary Statistics in college. I'm less than a week into class and I'm lost. Perhaps you could help.

Homework Statement


I have the answer to this as it's a practice question from the book. I don't know how to get it.
Consider the type of clothes dryer (gas or electric) purchased by each of five different customers at a certain store.

a.) If the probability that at most one of these purchases an electric dryer is .428, what is the probability that at least two purchase an electric dryer?

b.) If P(all five purchase gas) = .116 and P(all five purchase electric) = .005, what is the probability that at least one of each type is purchased?

Homework Equations



P(A') = 1 - P(A)
Others, maybe?

The Attempt at a Solution



a.) If P(at most one buys electric), then P(at least 2 buys electric) is 1-P(at most one buys electric). So, P(at least 2 buys electric) = .572, correct?

b.)
P(All 5 Gas) = .116
P(All 5 Elec) = .005

That's as far as I get, writing down the numbers. I have no idea where to go. In fact, I'm having an awful trouble with this material altogether. I just can't visualize any of it, or put what I'm trying to do into words easily.
 
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It's the same reasoning as the first one. Think about it. If it's NOT true that 5 gas or 5 electric were purchased, then at least one of each was purchased. So P(at least one of each)+P(all 5 gas)+P(all 5 electric)=1.
 
I don't understand your reasoning for assembling the equation that way. As I said, I'm having a really hard time equating the basic probability equations to the ideas in this chapter. I understand that the sample space = 1, but not the left side.
 
exitwound said:
I don't understand your reasoning for assembling the equation that way. As I said, I'm having a really hard time equating the basic probability equations to the ideas in this chapter. I understand that the sample space = 1, but not the left side.

It's logic. One and only one of the alternatives on the left can occur. So they cover the whole sample space. So their sum is 1. Take any sale pattern that can occur, think about which pattern it fits.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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