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I understand that all the possible outcomes are: HH, HT, TH, and TT.

The sample space is: S = {HH, HT, TH, TT}

But what are the possible events?

My textbook says that it is 2^n. So my understanding is that n is the number of outcomes in the sample space, which in my case is 4.

So 2^4 = 16.

That means that I should have 16 total events if I flip my single coin twice. How is this possible? As far as I'm concerned, if I flip a coin twice, I can only get two heads, two tails, or one head/one tail.

Regardless, from what I understand, this is what the answer should be:

{HH, HH}

{HH, HT}

{HH, TH}

{HH, TT}

{HT, HH}

{HT, HT}

{HT, TH}

{HT, TT}

{TH, HH}

{TH, HT}

{TH, TH}

{TH, TT}

{TT, HH}

{TT, HT}

{TT, TH}

{TT, TT}

This is 16 events... but my textbook also says that the null set is a subset of every set. If I add the null set, that would make it 17 events.

What am I missing?

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  • $\begingroup$ @MathMajor This is what my textbook says regarding null/empty sets: i.imgur.com/KsZEPdv.png $\endgroup$ – edgar_is Jul 14 '16 at 6:17
  • $\begingroup$ The empty set is an event, though not a basic event. One always needs the empty set in a definition of probability. The probability of the universal set is 1, and events are closed under complement (and union, intersection). $\endgroup$ – coffeemath Jul 14 '16 at 6:18
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    $\begingroup$ Your list is not right. It should be $\emptyset$, $\{HH\}$ and three others like it, $\{HH, HT\}$ and five others like it, $\{HH,HT,TH\}$ and three others like it, and finally $\{HH,HT,TH,TT\}$. $\endgroup$ – André Nicolas Jul 14 '16 at 6:18
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    $\begingroup$ @AndréNicolas Thanks for the hint, is this what you mean? { }, {HH}, {HT}, {TH}, {TT}, {HH, HT}, {HH, TH}, {HH, TT}, {HT, TH}, {HT, TT}, {TH, TT}, {HH, HT, TH}, {HH, HT, TT}, {HH, TH, TT}, {HT, TH, TT}, {HH, HT, TH, TT} $\endgroup$ – edgar_is Jul 14 '16 at 6:46
  • $\begingroup$ @edgr.sanchez: You are welcome. Yes, that's right. $\endgroup$ – André Nicolas Jul 14 '16 at 6:52
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Once the "basic events" are defined as some set $B,$ then by definition the events are the subsets $E$ of $B,$ including the empty subset. So for example you did not list any subsets of sizes one or three (or all four), and of course you're right that the empty set is technically an event. By the way in your list of size two events there are lots of multiple listings, since there should only be $\binom{4}{2}=6$ of size 4.

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    $\begingroup$ { }, {HH}, {HT}, {TH}, {TT}, {HH, HT}, {HH, TH}, {HH, TT}, {HT, TH}, {HT, TT}, {TH, TT}, {HH, HT, TH}, {HH, HT, TT}, {HH, TH, TT}, {HT, TH, TT}, {HH, HT, TH, TT} Ok, I think I got it. Is this correct? $\endgroup$ – edgar_is Jul 14 '16 at 6:42
  • $\begingroup$ @edgr.sanchez The list in your comment seems correct to me. $\endgroup$ – coffeemath Jul 14 '16 at 7:03

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