50 coin tosses, probability of 25 heads vs 3 heads

  • Thread starter quantum13
  • Start date
  • Tags
    Probability
In summary, the question asks for the probability of getting 25 heads out of 50 coin tosses, and also the probability of getting 3 heads out of 50 tosses. The equation to find this probability is P = (n choose k) * (0.5)^n, where n is the total number of tosses and k is the desired number of heads. The first probability can be calculated as (50 choose 25) * (0.5)^50, and the second as (50 choose 3) * (0.5)^50. The key point to note is that the order of the heads does not matter, so combinations, rather than permutations, should be used in the equation.
  • #1
quantum13
66
0

Homework Statement


Toss a coin fifty times. What is the probability of throwing 25 heads, and the probability of throwing 3 heads?


Homework Equations


P = 0.5 ^ n, n = total tosses
other equations?

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?
 
Physics news on Phys.org
  • #2
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)
 
  • #3
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?
 
  • #4
quantum13 said:

Homework Statement


Toss a coin fifty times. What is the probability of throwing 25 heads, and the probability of throwing 3 heads?


Homework Equations


P = 0.5 ^ n, n = total tosses
other equations?

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]
 
  • #5
HallsofIvy said:
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 :smile:

For example, let's 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.
 
  • #6
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.
 
  • #7
D H said:
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 :smile:

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. :smile:
 
  • #8
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
 

What is the probability of getting 25 heads out of 50 coin tosses?

The probability of getting exactly 25 heads in 50 coin tosses is 0.112455. This can be calculated using the binomial probability formula: P(x) = nCx * p^x * (1-p)^(n-x), where n is the number of trials (50 in this case), p is the probability of success (0.5 for a fair coin), and x is the number of successes (25 in this case).

What is the probability of getting exactly 3 heads out of 50 coin tosses?

The probability of getting exactly 3 heads in 50 coin tosses is 0.048131. This can also be calculated using the binomial probability formula.

What is the most likely outcome of 50 coin tosses?

The most likely outcome of 50 coin tosses is to get 25 heads. This is because the probability of getting 25 heads is the highest at 0.112455, compared to all other outcomes.

Can the probability of getting 3 heads out of 50 coin tosses ever be higher than the probability of getting 25 heads?

No, the probability of getting 3 heads out of 50 coin tosses can never be higher than the probability of getting 25 heads. This is because the probability of getting 25 heads is always higher than the probability of getting 3 heads, as calculated using the binomial probability formula.

How many heads or tails can we expect to get in 50 coin tosses?

On average, we can expect to get 25 heads and 25 tails in 50 coin tosses. This is because the expected value of a binomial distribution is given by E(x) = np, where n is the number of trials (50) and p is the probability of success (0.5 for a fair coin). Therefore, E(x) = 50 * 0.5 = 25.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
6
Views
2K
  • Programming and Computer Science
Replies
10
Views
993
  • Precalculus Mathematics Homework Help
Replies
7
Views
3K
  • Precalculus Mathematics Homework Help
Replies
5
Views
1K
  • Precalculus Mathematics Homework Help
Replies
16
Views
2K
  • Set Theory, Logic, Probability, Statistics
Replies
6
Views
2K
Replies
1
Views
1K
  • Precalculus Mathematics Homework Help
Replies
7
Views
1K
  • Precalculus Mathematics Homework Help
Replies
24
Views
3K
  • Set Theory, Logic, Probability, Statistics
Replies
14
Views
3K
Back
Top