Question

In an experiment, a fair coin is tossed 9 times and the face that appears is recorded. How many elements of the sample space will have no heads ? How many elements of the sample space will have exactly one head ? How many elements of the sample space will start or end (or both) with a head and have a total of exactly two heads ? How many elements of the sample space will start and end with a head with a total of exactly three heads ?

Answer 1

Please help :(

Answer 2

To have no heads, then you would have to have all tails. There is only one way to have all tails.

Answer 3

For one head, you would need to put that somewhere in the 9 positions. So there are 9 ways.

Answer 4

I think that is not correct

Answer 5

Oh, how so?

Answer 6

Hints:

p(No Heads) = \({9}\choose{0}\)\(\left(\dfrac{1}{2}\right)^{0}\left(\dfrac{1}{2}\right)^{9}\) = p(No Tails)

p(One Heads) = \({9}\choose{1}\)\(\left(\dfrac{1}{2}\right)^{1}\left(\dfrac{1}{2}\right)^{8}\) = p(One Tails)

p(Two Heads) = \({9}\choose{2}\)\(\left(\dfrac{1}{2}\right)^{2}\left(\dfrac{1}{2}\right)^{7}\) = p(Two Tails)

Answer 7

I dont have to find the probability, I need to find the number of elements in sample space

Answer 8

@nkaty Why do you doubt my solutions so far?

Answer 9

I dont know but it doesn't seems to be correct

Answer 10

What an odd thing to say. Once you have the probabilities, and the total number of possible outcomes, they are the same.

For example, \(2^{9} = 512\) -- This is the number of possible outcomes.
\(p(No\;Heads) = {9\choose{0}}\left(\dfrac{1}{2}\right)^{0}\left(\dfrac{1}{2}\right)^{9} = 9\cdot 1\cdot \dfrac{1}{512} = \dfrac{9}{512}\)

Thus: \(\dfrac{9}{512}\cdot 512 = 9\) -- Looks like what it IS has been shown to be more important than whatever is SEEMS.

Answer 11

Thanks @tkhunny but can you help with the last part too please? :)

Answer 12

Do you still doubt my method even though it has been shown to be correct?

Answer 13

Or at least lead to the same outcome?

Answer 14

I am sorry @wio, I got confused :( Sorry

Answer 15

For the final question, we split it up into three tasks:

First one is heads, and last one isn't
First one is heads and last one is heads.
First one isn't heads, and last one is.

Answer 16

1) First one is heads, last one isn't. We just need to put a single head among the remaining 7 spots, as the rest will be tails.
2) There is only one way to do this, first and last are heads and the rest are tails.
3) This is identical to (1). So there are seven ways.

We combine these ways by adding them.\[
7+1+7 = 15
\]

Answer 17

(9,0) is not 9P0 right?

Answer 18

It is a symmetric distribution. I have already given you six of the 10 probabilities. You should be able to answer many other questions from these probabilities.

How many will have exactly one heads: \(\dfrac{9}{512}\cdot 512 = 9\)

Correction of Previous error. Somehow, I believed for a typo moment that 9 Choose 0 wa s 9, rather than the correct 1.

How many will have exactly zero heads: \(1\cdot 1\cdot \dfrac{1}{512}\cdot 512 = 1\)

To @wio's important point, unique answers don't care how you find them. If you can reason it out without all the arithmetic, please do so.

My favorite example is the number of games required to pick a champion from a field of 64 teams in a single-elimination tournament. You can count all the games if you wish, but it's a LOT easier just to say, 64 teams, 1 champion, 63 losers, done. 63 games. No need for fancy arithmetic on this one.

Answer 19

Thank you @wio and @tkhunny

Answer 20

As for the final one....
We know the first and last have a head, so we just put a head among the remaining 7 in the middle. So it is just 7.

Answer 21

These questions did not even require any combinatorics. It's almost a shame.

Answer 22

How is (9 0 ) = 9? @tkhunny

Answer 23

Okay yes it is 1, I read in your next comment

Answer 24

Repeated For Convenience:

Correction of Previous error. Somehow, I believed for a typo moment that 9 Choose 0 was 9, rather than the correct 1.

Answer 25

:) thanks