flip 2 coins and roll a die

am I incorrect to think there are n(Ω) = 24?

/edit

omega = Ω = sample space = S

woops in title the question should be "find sample SPACE" not size

This post was edited by Regargar on Oct 4 2017 02:22am

24 is the correct number of outcomes if it is a 6-sided die.

If it wants the sample space detailed, you can always write out the genericized version to save space.

S = {(c1, c2, d) | c € {H, T}, d € {1, 2, 3, 4, 5, 6}}


Sorry for the notation. c1 and c2 should be subscripts. The Euro sign is meant to stand in for "is an element of". I don't have the epsilon handy on this phone keyboard.



This post was edited by timmayX on Oct 4 2017 11:20am

Thanks man I will ask my ta about this also

If you're not wanting to use notation (though the above posted is VERY useful set theory) you can get comfortable with the idea like this.

Imagine all possible outcomes. Ignoring the die roll:

HH, HT, TT, TH

Now we can easily imagine that after any of these 4 events you may roll a number 1 through 6 which would allow 6 possible distinct events after each of these 4. 4 events each with 6 possible endings.

4 x 6 =24

2x2x6, assuming permutation (order matters)

Might be mistaken in your desired answer, but would heads, tails, # be the same as tails, heads, #?

This variant consideration would generate results biased towards groups containing one heads and one tails.

End result would be 2 groups of 6 being as probable as the one group of 12... I am too tired to convert this to function form, bit I think you will see where I am going.