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paulb203

Member

Registered: 2023-02-24

Posts: 314

Systematic Listing / Product Rule for Counting

Bob is going to flip a coin 3 times

List all the possible outcomes

*

I did this 'systematically' as the topic heading suggested, and got HHH HHT etc (8 possible outcomes).

But what is the product rule (I think that's the correct term?) for working out the possible outcomes?

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Phrzby Phil

Member

From: Richmond, VA

Registered: 2022-03-29

Posts: 50

Re: Systematic Listing / Product Rule for Counting

The result of each flip is independent of the others.  Each flip has 2 equally likely outcomes - assuming a fair coin.

So 2^3 = 2x2x2 = 8 equally likely outcomes.

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paulb203

Member

Registered: 2023-02-24

Posts: 314

Re: Systematic Listing / Product Rule for Counting

Thanks, Phil.

But why does 2^3 give us our answer?

I get that there are 2 possible outcomes, and we flip 3 times, so I can see the 2 and the 3, but don't get why 2^3 works.

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Prioritise. Persevere. No pain, no gain.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: Systematic Listing / Product Rule for Counting

If the probability of event A is a, and event B is b then the probability of both occurring is a times b unless one event occurring effects and changes the probability for the other.

As the coin flips are independent it's 1/2 every time hence 1/2 x 1/2 x 1/2.

If the events are not independent then you have take account of the change in probability.

eg what's the probability of drawing two aces from a pack of 52 cards.

Drawing the first ace has probability 4/52 or 1/13.  But now there are only 51 cards amnd 3 aces so the probability of drawing another ace is 3/51 or 1/17

So to get two multipliy these probs to get 1/13 x 1/17 = 1/221  Note you still have to multiply the Ps.

Bob

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Phrzby Phil

Member

From: Richmond, VA

Registered: 2022-03-29

Posts: 50

Re: Systematic Listing / Product Rule for Counting

All possible outcomes:
First flip is either H or T - 2 events.
For each of these 2 events, second flip is either H or T - 2 events. So far, for 2 flips: HH, HT, TH, TT; count = 2*2.
For each of these 4 events, ditto; count = 4*2 = 2*2*2 = 2^3.

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World Peace Thru Frisbee

paulb203

Member

Registered: 2023-02-24

Posts: 314

Re: Systematic Listing / Product Rule for Counting

Thanks, Bob, thanks, Phil.

I think I've got it.

The possible outcomes for;

1 flip = 2^1 (H or T)

2 flips = 2^2 (HH, HT, TT, TH)

3 flips = 2^3 (HHH, etc)

and,

4 flips = 2^4 (16 possible outcomes; HHHH, etc)

5 flips = 2^5

etc, etc?

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Jai Ganesh

Administrator

Registered: 2005-06-28

Posts: 48,396

Re: Systematic Listing / Product Rule for Counting

Hi,

.

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