Counting problem, exactly 5 heads obtained, coin

In summary, there are 1024 possible outcomes of the coin-tossing experiment. Out of these, 979 contain at least one head.
  • #1
mr_coffee
1,629
1
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! = 11
Any help would be great, I'm having issues figuring out how to choose once i figure out the method it seems.
 
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  • #2
I am pretty sure (b) and (c) are wrong, but the rest are right.

One thing you could notice is that [tex]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}[/tex]

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).

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

So [tex]\binom{10}{3}[/tex] 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.
 
Last edited:
  • #3
Thanks matt i'll take another look at it once we get our homework back
 

Related to Counting problem, exactly 5 heads obtained, coin

1. How do you calculate the probability of exactly 5 heads obtained in a coin toss?

The probability of exactly 5 heads obtained in a coin toss can be calculated by using the binomial probability formula. This formula takes into account the number of trials (n), the probability of success (p), and the number of successes (k). In this case, n would be the number of coin tosses, p would be 0.5 (assuming a fair coin), and k would be 5. Plugging these values into the formula will give you the probability of getting exactly 5 heads in a coin toss.

2. What is the sample space for a coin toss when trying to obtain exactly 5 heads?

The sample space for a coin toss when trying to obtain exactly 5 heads would be all possible outcomes of 5 tosses, which would be HHHTT, HHTHT, HTHTH, THTHT, etc. Essentially, it would be all combinations of 5 heads and 5 tails.

3. Can you use the same formula to calculate the probability of obtaining a different number of heads in a coin toss?

Yes, the same formula for calculating the probability of exactly 5 heads can be used to calculate the probability of obtaining a different number of heads. You would just need to plug in the appropriate values of n, p, and k based on the desired number of heads.

4. Is the probability of obtaining exactly 5 heads in a coin toss affected by the previous outcomes?

No, the probability of obtaining exactly 5 heads in a coin toss is not affected by the previous outcomes. Each coin toss is an independent event and the probability of getting 5 heads remains the same regardless of the previous outcomes.

5. How does the number of coin tosses affect the probability of obtaining exactly 5 heads?

The number of coin tosses does not affect the probability of obtaining exactly 5 heads. As long as the coin is fair, the probability of getting 5 heads remains the same regardless of the number of tosses. However, the probability of getting exactly 5 heads may increase with more tosses since there are more opportunities for it to occur.

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