1. ## Difficulty Understanding Something...

Hi everybody,

If a problem were asked, for example:

10 coins are flipped. What is the probability that 10 are heads?

Naturally, I would say, the probability of one coin being heads is .5, so, the probably of 8 coins being heads would be .5 * .5 * .5 * .5 * .5 * .5 * .5 * .5 or .5^8.

That's just what makes sense to me, and equals .0039.

However, the book goes about this question a very strange way. In my chapter on Binomial distribution, it uses combinations and says:

10 C 8 * (.5)^8 * (.5)^2

This method yields .043, which is supposedly the correct answer. These two methods are different and their answers are different. This entire chapter on binomial distribution is very bizare to me and makes no logical sense to me.

If anybody can help me understand why on earth this makes sense, I would appreciate it.

Thanks!
Ethan

2. ## Maybe

Originally Posted by EthanDavis
Hi everybody,

If a problem were asked, for example:

10 coins are flipped. What is the probability that 10 are heads?

Naturally, I would say, the probability of one coin being heads is .5, so, the probably of 8 coins being heads would be .5 * .5 * .5 * .5 * .5 * .5 * .5 * .5 or .5^8.

That's just what makes sense to me, and equals .0039.

However, the book goes about this question a very strange way. In my chapter on Binomial distribution, it uses combinations and says:

10 C 8 * (.5)^8 * (.5)^2

This method yields .043, which is supposedly the correct answer. Although the two answers are VERY close, they are different. This entire chapter on binomial distribution is very bizare to me and makes no logical sense to me.

If anybody can help me understand why on earth this makes sense, I would appreciate it.

Thanks!
Ethan
Maybe J am misinterpreting the questino...I didnt really read the whole thing haha...but the probability of getting all heads with $n$ flips with a fair coin is $\bigg(\frac{1}{2}\bigg)^{n}$

3. Originally Posted by EthanDavis
Hi everybody,

If a problem were asked, for example:

10 coins are flipped. What is the probability that 10 are heads?

Naturally, I would say, the probability of one coin being heads is .5, so, the probably of 8 coins being heads would be .5 * .5 * .5 * .5 * .5 * .5 * .5 * .5 or .5^8.

That's just what makes sense to me, and equals .0039.

However, the book goes about this question a very strange way. In my chapter on Binomial distribution, it uses combinations and says:

10 C 8 * (.5)^8 * (.5)^2

This method yields .043, which is supposedly the correct answer. Although the two answers are VERY close, they are different. This entire chapter on binomial distribution is very bizare to me and makes no logical sense to me.

If anybody can help me understand why on earth this makes sense, I would appreciate it.

Thanks!
Ethan
First learn to read the question. From what you write the real question appears to be: You flip 10 coins, what is the probability that they land with exactly 8 heads uppermost.

Now this is a binomial distribution question and the answer is b(8;10,0.5), which is the answer you quote.

(try direct enumeration of all the equally likely cases with say 3 coins and 2 heads showing to see what is going on)

RonL

4. I can read just fine, thank you. And the question I quoted was just simplified; put in my own words. Sorry if they're not worded exactly as they're supposed to.

Anyway, you told me that I'm supposed to use binomial distribution. Thanks. For someone criticizing on reading skills, you didn't notice that I wanted to know WHY to use binomial distribution, rather than the method that made sense to me.

As you see, Mathstud28 agreed with me. He also thought that it should be .5^8. So I am not the only one confused here.

So again, if anybody can explain to me the use of .5^n (what makes the most sense to me) vs. combinations/binomial distribution (which doesn't make any sense to me), please explain!

If you can help me figure this out, I would be oh so grateful.

Thanks again!
Ethan

5. Originally Posted by EthanDavis
I can read just fine, thank you. And the question I quoted was just simplified; put in my own words. Sorry if they're not exact to your liking.

Anyway, you told me that I'm supposed to use binomial distribution. Thanks. For someone criticizing on reading skills, you didn't notice that I wanted to know WHY to use binomial distribution, rather than the method that made sense to me.

As you see, Mathstud28 agreed with me. He also thought that it should be .5^8. So I am not the only one confused here.

So again, if anybody can explain to me the use of .5^n (what makes the most sense to me) vs. combinations/binomial distribution (which doesn't make any sense to me), please explain!

If you can help me figure this out, I would be oh so grateful.

Thanks again!
Ethan
Hello Ethan

I understand your frustration but in short you are asking someone to teach you a whole topic online, It would take a member of this forum a considerable amount of time to type up notes on whole topic for you.

To makes things easier, can you tell us what do you know. Are you familiar with discrete probability distributions ? it is important that you know of them before you can do this.

Also take CB suggestion.
(try direct enumeration of all the equally likely cases with say 3 coins and 2 heads showing to see what is going on)
you could get

$H , H , T$
$H , T , H$
$T , H , H$

so there are 3 possible way in which the event of getting 2 heads after tossing three coins can occur. so the probability of this event happening is not $\left( \frac{1}{2} \right)^3$ it is three times it because there are three ways in which the even can occur.

are you starting to understand where the combinations formula fits into this?

I hope this helps

Bobak

6. Originally Posted by EthanDavis
If a problem were asked, for example:
10 coins are flipped. What is the probability that 10 are heads?
I suspect that you have a typo. The 'red' 10 should be 8.
In other words you have a string "HHHHTTHHHH".
But do you see that the two T's can happen is many places: $\binom {10}{2}=45$
Any string has a probability of $\left ( \frac {1}{2} \right )^{10}$

7. Originally Posted by EthanDavis
As you see, Mathstud28 agreed with me. He also thought that it should be .5^8. So I am not the only one confused here.
I'm afraid he didn't, read what he wrote:

"but the probability of getting all heads with n flips with a fair coin is (1/2)^n"

You see he is refering to what you did ask, and allowing for the uncertainty as to whether the number of flips is 8 or 10. You see his confusion is about what your question is.

RonL

8. Originally Posted by EthanDavis
I can read just fine, thank you. And the question I quoted was just simplified; put in my own words. Sorry if they're not worded exactly as they're supposed to.

Anyway, you told me that I'm supposed to use binomial distribution. Thanks. For someone criticizing on reading skills, you didn't notice that I wanted to know WHY to use binomial distribution, rather than the method that made sense to me.
You have a text book with a section on the binomial distribution. With a very high probability it explains why the binomial distribution is the appropriate distribution for the number of successes in a set of Bernoulli trials.

Also you are sitting in front of a computer and so have access to search engins into which you can type "binomial distribution" and a large proportion of the hits will give an explanation as good or better than the one I would give.

The attachment shows what Wikipedia (the first hit from a Google search for "binomial distribution") has to say.

RonL

9. Thanks for the replies, and I apologize for slipping up my question.

I have found my answer in the fact that Binomial Distribution accomidates for results that have ALREADY been found, while the exponential equation (.5^n) is used to PREDICT results.

Thanks,
Ethan

10. ## 10 coin flips-probalbility of 2 tails 8 heads

originally posted by ethan davis. answered by others p= 0.044
in the hope of clarifying the answer for ed here is another approach.
when a coinis flipped there are two outcomes.on the next flip two more outcomes making 2x2=4 outcomes. next 2x2x2=8outcomes. after 10 flips the outcomes total 2E10 or 1024. these are all the possible outcomes.
for exactly 2 tails these can show up on only two of the 10 flips(1-10) but which number flips.so the question is how many two flip numbers can be generated from the ten.this is the number of two nunber combinations from the ten or (10) =45 these are favorable
2

probability_favorable outcomes 45/2E10=0.044
total outcomes