Question 488990
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If I drop three coins on a table, what is the probability of finding
1. exactly 2 heads
2. at most 1 head
3. at least 2 heads
4. at least 1 head
sample space = 8
I came up with HHT, HTH, THH,HTT,THT,TTH,TTT,HHH.
state the probability of each in the form of fraction and percent.
For 1,2,3 I came up with 1/4 or .25, and for 4 = 1/6.
Have I done this correctly, or can you show me a better way to show this??
Is this a form of "classical probability"??
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        As I read the problem and your solution in your post,

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;it becomes clear to me that your level is &nbsp;<U>introductory</U> &nbsp;in &nbsp;Probability.


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;At this level, &nbsp;the problem assumes that you will prepare the space of events

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;and will answer questions based on this table, &nbsp;without using the combinatorics terms/formulas

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;of &nbsp;Binomial distribution, as tutor &nbsp;@Theo does.


&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;So, &nbsp;I &nbsp;will not demonstrate how smart &nbsp;I &nbsp;am, &nbsp;but simply will prepare the table of events

&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;and will show you how to operate with it.



<pre>
The events are in the second column of this table below.

It has 8 elements, so the space of events has 8 elements with associated probability 
1/8 for each event.


           Q1    Q2    Q3    Q4

  1  HHH                *     *

  2  HHT    *           *     *

  3  HTH    *           *     *

  4  HTT          *           *

  5  THT          *           *

  6  TTH          *           *

  7  TTT          *         

  8  THH    *           *     *



For Question 1, I marked by (*) 3 appropriate events in column Q1.

                So, the probability P(Q1) = 3/8.



For Question 2, I marked by (*) 4 appropriate events in column Q2.

                As a reminder, "at most 1 head" means 0 or 1 heads.

                So, the probability P(Q2) = 4/8 = 1/2.



For Question 3, I marked by (*) 4 appropriate events in column Q3.

                As a reminder, "at least 2 heads" means 2 or 3 heads.

                So, the probability P(Q3) = 4/8 = 1/2.



For Question 4, I marked by (*) 7 appropriate events in column Q4.

                As a reminder, "at least 1 head" means 1 or 2 or 3 heads.

                So, the probability P(Q4) = 7/8.
</pre>

Solved. &nbsp;&nbsp;All questions are answered.


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So, &nbsp;at this level &nbsp;(which is &nbsp;YOUR &nbsp;level), &nbsp;it is &nbsp;HOW &nbsp;TO &nbsp;you should analyze this problem,

how to you should solve this problem and how to you should present your solution.


It is what your teacher does expect from you.