What is the Probability of Getting More Heads than Tails in 4 Coin Tosses?

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Discussion Overview

The discussion revolves around calculating the probability of obtaining more heads than tails in four tosses of a coin. It includes various approaches to the problem, including the use of the binomial distribution and combinatorial reasoning.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant asks about the probability of getting more heads than tails when a coin is tossed four times.
  • Another participant suggests using the binomial distribution to solve the problem.
  • A different participant estimates the probability to be about 25% without providing a detailed rationale.
  • One participant proposes an alternative method by considering specific combinations of heads and tails that result in more heads than tails.
  • Another participant questions whether there is a formula for the probability when tossing the coin four times, mentioning "4 choose 2" in relation to the outcomes.
  • A participant provides a detailed breakdown of the probabilities for various outcomes when tossing the coin multiple times, emphasizing the need to count combinations that yield more heads than tails.
  • One participant expresses gratitude for the assistance received, indicating that the discussion has been helpful.

Areas of Agreement / Disagreement

There is no consensus on the exact probability or the method to calculate it, as participants propose different approaches and estimates. The discussion remains unresolved regarding the final probability value.

Contextual Notes

Participants reference the binomial distribution and combinatorial methods without fully resolving the mathematical steps or assumptions involved in their calculations.

tae3001
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A coin is tossed 4 times. What is the probability of getting more heads than tails? :confused:
 
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Use the binomial distribution.
 
I'm guessing about 25% :confused:
 
Juvenal's suggestion is good, but I thought I'd give you an alternate way of solving the problem. What is the probability of a certain combination och heads and tails? For example: first throw you ger tail, the three following throws heads come up. Then, think about in how many such combinations you would get more heads than tails.
 
juvenal said:
Use the binomial distribution.



is there.. a formula for it..
if its tossed 4 times.. and there are 2 sides..
wouldn't it be 4 choose 2?
if there are more H that t would it be HHH and T...
i hate probability questions.. :eek:
 
Every time you throw there is 50% chance of T and 50% of H.
So, if you throw once you have (P(x)=probability that x happens):
1. P(H)=1/2
2. P(T)=1/2.
If you thow twice:
1. P(H,H)=(1/2)*(1/2)=1/4
2. P(H,T)=(1/2)*(1/2)=1/4
3. P(T,H)=(1/2)*(1/2)=1/4
4. P(T,T)=(1/2)*(1/2)=1/4
If you throw four times there is 1/16 chance you'll get one of the 16 combination:
1. HHHH, 2. HHHT, 3. HHTH,... etc.
These three examples are all more Heads than Tails, how many more such combinations are there? If you can answer that, you're done.
 
I get it.. thanx.. u guys helped me graduate..(LOL).. awesome..
 

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