Welcome to our community

Be a part of something great, join today!

[SOLVED] another coin toss problem

karush

Well-known member
Jan 31, 2012
2,719
Tim carried out an experiment where he tossed 20 coins together and observed the number of heads showing. He repeated this experiment 10 times and got the following results

11,9,10,8,13,9,6,7,10,11

a) Use Tim's data to get the probability of obtaining a head.
the ans to this was .47 but I don't know how that was derived

b) He tossed the 20 coins for the 11th time, How many heads should he expect to get?
I would guess 9 but not sure what the mathematical calc would be

c) He tossed the coins 1000tmes. How many heads should he expect to see.
since this a much longer data base I would think this would smooth out to
50% or 500 heads
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
a) add up the heads, and divide this by the number of times a coin was tossed (number of coins times number of tosses). What do you get?

b) multiply the result from part a) times the number of coins, and round to the nearest integer.

c) He is tossing 20 coins 1000 times, so there are 20,000 outcomes. What do you think you need to do with this number?
 

Bacterius

Well-known member
MHB Math Helper
Jan 26, 2012
644
Are you familiar with the binomial distribution? Questions 1 and 2 might require its use unless I've misinterpreted the problem (damn statistics, arr)

The last question doesn't - the probability of one coin being heads is 1/2, so the expected value of 20 * 1000 coin flips is 1/2 * 20000 = 10000 heads (and thus 10000 tails). Or, you can also use the fact that the binomial distribution is symmetrical for p = 1/2.
 

karush

Well-known member
Jan 31, 2012
2,719
Are you familiar with the binomial distribution?

Um no, statistics is a very primitive subject to me, but need to learn it:cool:
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
You don't need the binomial distribution here...if you had been asked what is the probability of getting a certain number (or range) of heads, then you would need it.
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,042
You don't need the binomial distribution here...if you had been asked what is the probability of getting a certain number (or range) of heads, then you would need it.
I would like to ask for clarification, if you would be so kind :) I'm interpreting what you wrote as you don't need to use the binomial distribution here to get the correct answer but you could if you wanted to. Is that correct?

The design is set up where each trial has 20 flips so this isn't the typical binomial distribution situation, but since since we're just counting the total number of heads and they flips are independent we can say this experiment is analogous to flipping each coin individually.

If every one agrees with that then we have a random variable, $H$, which represents the number of heads, then it's a binomially distributed random variable or \(\displaystyle \text{Bi}(n,.5)\) and it's expected value after $n$ trials is $E[H]=np$ or $E[H]=.5n$. For $n=20000$ we have $E[H]=(.5)(20000)=10000$.

Now this is exactly what everyone said but to put it in the context of a binomial distribution might be overkill for karush, especially if this isn't how the class is being taught. I just wanted to comment on what I think Bacterius was going for. :)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I would like to ask for clarification, if you would be so kind :) I'm interpreting what you wrote as you don't need to use the binomial distribution here to get the correct answer but you could if you wanted to. Is that correct?...
I don't know how I missed this...yes, that is exactly what I meant. :D