-aux11.outcomes in card drawing

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Discussion Overview

The discussion revolves around the outcomes of drawing two cards from a standard deck of 52 cards. Participants explore the sample space and the calculation of the total number of outcomes for this experiment, focusing on combinatorial reasoning.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant lists a sample space for drawing two cards but does not clarify the notation used.
  • Another participant introduces the concept of combinations, suggesting that the order of drawing cards does not matter, and proposes using the combination formula \binom{52}{2} to calculate outcomes.
  • A third participant confirms the calculation of C(52,2) as 1326 and references the formula for combinations, expressing confusion about its absence in the textbook.
  • A fourth participant agrees with the formula and provides an alternative explanation using a step-by-step approach to arrive at the same outcome, emphasizing the division by 2 to account for the order not mattering.

Areas of Agreement / Disagreement

Participants generally agree on the calculation of the number of outcomes as 1326 using combinations, but there is uncertainty regarding the absence of this method in the textbook and the notation used in the sample space.

Contextual Notes

There are limitations regarding the clarity of the sample space notation and the assumptions about the methods expected in the textbook, which remain unresolved.

karush
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We draw cards from a deck of 52 cards

a) list the sample space if we draw two cards at a time

my ans to this is ((H2,S2)...(C2,D2))

b) How many outcomes do you have in this experiment

but I don't know how they got the answer of 1326

thanks ahead...
 
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Are you familiar with combinations? It's a way of counting groupings when the order doesn't matter. Since for this experiment AhAs is the same as AsAh, then we can say the order doesn't matter and use combinations. So the total number of outcomes is [math]\binom{52}{2}[/math]. Do you know how to calculate that?
 
my calculator gave

C(52,2) = 1326

so I looked it up $C(n,r) = \frac{n!}{(n-r)!r!}$

However they didn't have this in the text?
 
Yep, that's the correct formula. I don't know why this isn't in your book. Perhaps they want you to use a more intuitive method. For the first card you have 52 choices and for the second you have 51 choices. However, since order doesn't matter you should divide by 2 since we don't want to count AsAh and AhAs twice. So you have [math]\frac{52*51}{2}=1326[/math]. That's exactly the same calculation as [math]\binom{52}{2}[/math] but approaches it from a different way in the set up.
 
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