Counting problem, exactly 5 heads obtained, coin

[mr_coffee]

Hello everyone, I'm having some issues on this problem:
A coin is tossed ten times. In each case the outcome H (for heads) or T (for tails) is recorded. (One possible outcome of the ten tossings is denoted THHTTTHTTH.)

I got a, d right i believe.
but I need someone to check if i did the others ones right.a. what is the total number of possible outcomes of the coin-tossing experiemnt?

I said: 2^10 = 1,024

b. In how many of the possible outcomes are exactly 5 heads obtained?
[# of outcomes with exactly 5 heads] = [total # of outcomes] - (10 choose 5)

This one I'm not sure what I'm suppose to do, when i say 10 choose 5, i mean you have a set of 10, and I'm wanting to choose 5 elements out of that set that are H but I'm not sure how to represeent this...

c. In how many of the possible outcomes is at least 8 heads obtained?
For this one would i do:
[# of outcomes with at least 8 heads obtained] = [total number of outcomes] - [number of outcomes that contain 8 heads]

= 1024 - (10 choose 8)
= 1024 - 10!/[8!(2!)]
= 1024 - 45 = 979

d. In how many of the possible outcomes is at least one head obtained?
[# of outcomes w/ at least 1 head] = [total # of outcomes] - [# of outcomes with no heads] = 1024 - 1 = 1023.

e. In how many of the possible outcomes is at most one head obtained?

I think i got this one, but the # of outcomes with 1 head i might have got wrong.

[# of outcomes with at most 1 head] = [# of outcomes with no heads] + [# outcomes with 1 head]
= 1 + (10 choose 1)
= 1 + 10!/[1!(10-1)!]
= 1 + 10!/9! = 11Any help would be great, I'm having issues figuring out how to choose once i figure out the method it seems.

 

[mattmns]

I am pretty sure (b) and (c) are wrong, but the rest are right.

One thing you could notice is that 1024 = 2^{10} = \binom{10}{0} + \binom{10}{1} + \binom{10}{2} + \binom{10}{3} + \binom{10}{4} + \binom{10}{5} + \binom{10}{6} + \binom{10}{7} + \binom{10}{8} + \binom{10}{9} + \binom{10}{10}

What this means is that you can partition the 1024 outcomes according to how many heads there are in each possible outcome (you could also replace heads with tails here).

\binom{10}{i} would be the number of outcomes with exactly i heads.

So \binom{10}{3} would be the number of outcomes with exactly 3 heads.

So you should be able to figure out (b) from this.

In your explanation you keep subtracting from 1024, I am not sure why you are doing this (well I have an idea why) but you should not be doing this. Remember X - A = A' (Where A,A' are subsets of X, and A' is the complement of A). That is, 1024 - number of sets with 5 heads = number of sets that do not have 5 heads.

In part (c) you are doing [total] - [number that have exactly 8], as I said above this would be the [number that do not have 8], not the number that have at least 8. The negation of [at least 8] is [less than 8], so you would have to say [number with at least 8] = [total] - [number with less than 8] but you would not want to do it this way since you would have more things to compute than if you just do it directly. So, I would just compute each case directly (that is, the number of outcomes for 8 heads, 9 heads, and 10 heads, and add these).

edit.. the LaTeX is not working, just click the "images" to see the code I guess.

 

[mr_coffee]

Thanks matt i'll take another look at it once we get our homework back