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50 coin tosses, probability of 25 heads vs 3 heads

by quantum13
Tags: coin, heads, probability, tosses
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quantum13
#1
Sep1-09, 04:37 PM
P: 66
1. The problem statement, all variables and given/known data
Toss a coin fifty times. What is the probability of throwing 25 heads, and the probability of throwing 3 heads?


2. Relevant equations
P = 0.5 ^ n, n = total tosses
other equations???

3. The attempt at a solution
the probability for each should be same, as in 0.5^50, for each, should it not?

when i asked this question in class, my statistics teacher gave me a weird look and moved on with the class. am i missing a fundamental point in logic or something more complex?
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rock.freak667
#2
Sep1-09, 05:22 PM
HW Helper
P: 6,202
H=head

P(H)=0.5

P(HH)=P(H)*P(H) (this is because they are independent events. So that getting a head on the first try does not influence the outcome of the second)
njama
#3
Sep1-09, 05:47 PM
P: 220
Ok, so you throw the coin 50 times. That is clear.

My first question, must that 25 heads be in row?

Also, do you know what probability is?

P=m/n , m - the number of possibilities that might occur, n - total number of possibilities

Also [itex]0 \leq P \leq 1[/itex].

In addition, just imagine if you throw the coin once, what is the probability that head or tail will be thrown?

What is the probability for 50 times?

HallsofIvy
#4
Sep1-09, 06:45 PM
Math
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P: 39,565
50 coin tosses, probability of 25 heads vs 3 heads

Quote Quote by quantum13 View Post
1. The problem statement, all variables and given/known data
Toss a coin fifty times. What is the probability of throwing 25 heads, and the probability of throwing 3 heads?


2. Relevant equations
P = 0.5 ^ n, n = total tosses
other equations???

3. The attempt at a solution
the probability for each should be same, as in 0.5^50, for each, should it not?

when i asked this question in class, my statistics teacher gave me a weird look and moved on with the class. am i missing a fundamental point in logic or something more complex?
I can't speak for why your teacher gave you a weird look but the point you are missing is that [itex](0.5)^{50}[/itex] is the probability of getting 25 heads and 25 tails in a specific order- say 25 heads followed by 25 tails or alternating heads and tails. There are
[tex]\left(\begin{array}{c}50 \\ 25 \end{array}\right)[/tex]
ways of ordering 25 heads and tails so the probability of getting 25 heads and 25 tails in any order is
[tex]\left(\begin{array}{c}50 \\ 25 \end{array}\right)(.5)^{50}[/tex].

Similarly, there are
[tex]\left(\begin{array}{c}50 \\ 3 \end{array}\right)[/tex]
ways of arranging 3 heads and 47 tails so the probability of flipping 3 heads and 47 tails in any order is
[tex]\left(\begin{array}{c}50 \\ 3 \end{array}\right)(.5)^{50}[/tex]
njama
#5
Sep2-09, 04:09 PM
P: 220
Quote Quote by HallsofIvy View Post
I can't speak for why your teacher gave you a weird look but the point you are missing is that [itex](0.5)^{50}[/itex] is the probability of getting 25 heads and 25 tails in a specific order- say 25 heads followed by 25 tails or alternating heads and tails. There are
[tex]\left(\begin{array}{c}50 \\ 25 \end{array}\right)[/tex]
ways of ordering 25 heads and tails so the probability of getting 25 heads and 25 tails in any order is
[tex]\left(\begin{array}{c}50 \\ 25 \end{array}\right)(.5)^{50}[/tex].

Similarly, there are
[tex]\left(\begin{array}{c}50 \\ 3 \end{array}\right)[/tex]
ways of arranging 3 heads and 47 tails so the probability of flipping 3 heads and 47 tails in any order is
[tex]\left(\begin{array}{c}50 \\ 3 \end{array}\right)(.5)^{50}[/tex]
The question is rather tricky, because it do not state the following:
-are those 25 heads thrown in a row?

-are there at least 25 heads and are they ordered?

I believe its the first one because the second one is too complicated to solve. In this case combinations work as same as permutations with repetition.

In cases like this one, I always take smaller values and find the pattern

For example, lets imagine that instead of 50 there are 5 throws. We need 3 heads (not ordered) in each 5 throws. So, the string will look like HHHTT , H - head ; T - tail.

Instead of permutations, we need permutations with repetition, so:

[tex]P_{(3,2)}5=\frac{5!}{2!3!}=10[/tex]

And the probability would be 10/25.

Do the same for the original problem.
D H
#6
Sep2-09, 04:33 PM
Mentor
P: 15,170
You are reading too much into the problem, njama. These problems are obviously asking about combinations, not permutations. If the questions were to find the probability of getting at least 25 heads (or 3) out of 50, they would have said just that.

Looking for patterns works for a small number of objects. This technique breaks down quickly as n increases.
njama
#7
Sep2-09, 05:28 PM
P: 220
Quote Quote by D H View Post
You are reading too much into the problem, njama. These problems are obviously asking about combinations, not permutations. If the questions were to find the probability of getting at least 25 heads (or 3) out of 50, they would have said just that.

Looking for patterns works for a small number of objects. This technique breaks down quickly as n increases.
@D H, permutations with repetition are the ones for this problem

m - permutations with repetition

n - variations with repetition.

P=m/n=P(25,25)50/250

The technique of doing the same task with smaller amounts, let you analyze the problem and give you an idea how to solve it. Nothing more.
Elucidus
#8
Sep2-09, 10:31 PM
P: 286
Minor point of clarification: A permutation with repetition of objects of exactly two kinds (like heads and tails) is a combination. They are the same thing.

--Elucidus


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