MHB -aux.18.A fair coin is tossed 8 times.

[karush]

1142

I know this a very easy problem, but trying to learn the probability formula's

(a) $P(H,H,H,H)= \frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}
=\frac{4}{2}=4$

or is there another better way to do especially if the problem is much bigger and more complicated.

 

[MarkFL]

You should observe that when you obtain a probability that is not on $[0,1]$ then you have made an error. A probability can range from 0 (impossible) to 1 (certain).

For this problem, you want to apply the binomial probability formula:

$$P(x)={n \choose x}p^x(1-p)^{n-x}$$

where:

$n$ is the number of trials

$x$ is the number of successes

$p$ is the probability of success for each trial

Can you identify the value of each variable for part a)?

 

[karush]

You should observe that when you obtain a probability that is not on $[0,1]$ then you have made an error. A probability can range from 0 (impossible) to 1 (certain).

For this problem, you want to apply the binomial probability formula:

$$P(x)={n \choose x}p^x(1-p)^{n-x}$$

where:

$n$ is the number of trials

$x$ is the number of successes

$p$ is the probability of success for each trial

Can you identify the value of each variable for part a)?

will try..

$\displaystyle P(4)={8 \choose 4}.5^4(1-.5)^{8-4}$

this is $= \frac{35}{128}$ ??

 

[Plato2]

will try..

$\displaystyle P(4)={8 \choose 4}.5^4(1-.5)^{8-4}$

this is $= \frac{35}{128}$ ??

You can write simply as $\dbinom{8}{4}(0.5)^{8}$.

So c) becomes $\sum\limits_{k = 3}^5 {\dbinom{8}{k}{{(0.5)}^{ 8}}} $

 

[karush]

You can write simply as $\dbinom{8}{4}(0.5)^{8}$.

So c) becomes $\sum\limits_{k = 3}^5 {\dbinom{8}{k}{{(0.5)}^{ 8}}} $

i got $46592$ is it really that large?

 

[Jameson]

No. That should be a positive 8 for the exponent. Like MarkFL said, all probabilities must be between 0 and 1. If you get anything outside of that range then you've made a mistake.

 

[karush]

No. That should be a positive 8 for the exponent. Like MarkFL said, all probabilities must be between 0 and 1. If you get anything outside of that range then you've made a mistake.

ok, corrected, it ans should be then $0.710938$

so should (b) b in W|A
Binom[8,3]*.5^8=0.21875

 

[Jameson]

ok, corrected, it ans should be then $0.710938$

You were right the first time.

 

[MarkFL]

I wanted to say a few words about the binomial probability formula. To see why the binomial coefficient is required, consider the following:

We are going to flip a coin $n$ times and we wish to find the probability that we get $m$ heads, where $0\le m\le n$. So let's draw $n$ place holders to represent the $n$ trials, or flips of the coin:

__, __, __, ..., __ (there are $n$ placeholders)

Now, for the first head, we have $n$ choices where we may place it. For the second head, we have $n-1$ choices, and so forth all the way down to $n-m+1$ for the last or $m$th head. Using the fundamental counting principle, we can calculate that there are:

$$n(n-1)(n-2)\cdots(n-m+1)=\frac{n!}{(n-m)!}$$

ways to make our choice. But, we have to consider that there are $m!$ ways to make each choice, since the order in which the same sequence of heads and tails is chosen does not matter, the result is the same. For example, suppose we are flipping the coin 3 times and we wish to calculate the probability of getting two heads.

So, as one possible outcome, we choose first the third trial to be a heads, and then the first trial to be a heads. This is identical to having chosen first the first trial to be heads and then the third trial to be heads. Both choices results in heads, tails, heads...they are identical and indistinguishable outcomes. So we need to divide by the number of ways to choose the same outcome, which is the factorial of the number of favorable outcomes. In other words, there are $m!$ ways to order $m$ objects.

So we may conclude that there are:

$$\frac{n!}{m!(n-m)!}={n \choose m}$$ ways to choose $m$ from $n$.

This is where the binomial coefficient comes from in the binomial probability formula. Then we take the product of all the probabilities, as the other factors in the formula, $m$ favorable and $n-m$ unfavorable. For each trial, where there are two possible outcomes, it is certain that one or the other outcome will happen. Suppose that $p$ is the probability for a favorable outcome for each trial. Let $q$ be the probability of the other or unfavorable outcome. Mathematically, we may state the certainty of either outcome occurring as:

$$p+q=1\,\therefore\,q=1-p$$

Now, putting everything together, we find that the probability of obtaining $m$ favorable outcomes during $n$ trials is:

$$P(m)={n \choose m}p^m(1-p)^{n-m}$$

Something we can sometimes use to simplify our calculations is the symmetry of the binomial coefficient, that is, the identity:

$${n \choose r}={n \choose n-r}$$

Do you see how you can use this in part c) of the given problem? Can you use the definition above of the binomial coefficient to verify this identity?

We should expect to find that the sum of all the possible probabilities is 1, and indeed we see (using the binomial theorem):

$$\sum_{m=0}^n{n \choose m}p^m(1-p)^{n-m}=(p+(1-p))^n=1^n=1$$

I hope this helps, as I find that when I understand where a formula comes from, it is much easier to remember and use. :D