# Using summation with probability question

1. Mar 30, 2005

### spikemurphy

Help With Probability Question

i have been working on this for a week can anyone help?

NOTE: Look at reply from moodoo for proper matematical symbols!!

I need the probability of being dealt a bridge hand with at least 5 hearts.

I have to possible answers but I have never done this before and dont know the proper way to type them but here goes...

1.)
summation sign with 13 on the top and k=5 on bottom x (C 13,k)(C 13,13-K)/C 52,13

2)same summation as above x C 13,k (.25)to the power of k(.75)to the power of 13-k/C 52,13

HELP WITH PROBABILITY QUESTION

Last edited: Mar 30, 2005
2. Mar 30, 2005

### Moo Of Doom

Yikes, that's some messy notation you got there...
Let me clean that up for you...

1. $$\sum_{k=5}^{13}{\frac{\left(\begin{array}{cc}13\\k\end{array}\right)\left(\begin{array}{cc}13\\13-k\end{array}\right)}{\left(\begin{array}{cc}52\\13\end{array}\right)}}$$

2. $$\sum_{k=5}^{13}{\frac{\left(\begin{array}{cc}13\\k\end{array}\right)*.25^{k}*.75^{13-k}}{\left(\begin{array}{cc}52\\13\end{array}\right)}}$$

Although I am a bit shaky on where your delimeters are. Tell me if any of the above aren't what you had in mind.

EDIT: Maybe that's more like it?

Last edited: Mar 30, 2005
3. Mar 30, 2005

### spikemurphy

summation

Yes, that is exactly what I meant. You are amazing. But is this the right answer for my question? Are both of these binomial formulas and is it ok to use a binomial formula with cards?

Thanks so much for your time>

Last edited: Mar 30, 2005
4. Mar 30, 2005

### xanthym

First, Binomial probabilities can only be used for INDEPENDENT events for which the probability of occurrence (or non-occurrence) of any event does NOT influence that for any other. Unfortunately, for a card deck, when you draw any one card you automatically change the probability of drawing the next (since you do not replace the drawn card). Thus, your solution #2 does not apply. (Binomial probabilities could be used, however, to determine the probability of at least 5 Tails when tossing 52 Fair Coins since each Coin is independent of the others.)

Your solution #1 is closer, but is still not proper. You are correct that there are $$\mathbb{C}_{13}^{52}$$ total ways of drawing 13 (combinations of) cards from the deck's 52. You must now determine how many of those contain at least 5 Hearts.

Let's start with exactly 5 Hearts. There are $$\mathbb{C}_{5}^{13}$$ ways of drawing those 5 from the deck's 13 Hearts. Further, there are $$\mathbb{C}_{8}^{39}$$ ways of drawing the 8 additional cards required for the hand from the remaining 39 deck cards (which are not hearts). Thus:

{Number Combinations of 5 Hearts in 13 Cards from 52 Card Deck} = $$\mathbb{C}_{(5)}^{(13)} \cdot \mathbb{C}_{(8)}^{(39)}$$

Similar reasoning can be applied to {5, 6, 7, ... , 12, 13 Hearts}, so the total number of combinations containing AT LEAST 5 Hearts is given by:

$$(1) \ \ \ \ \ \sum_{n=5}^{13} \mathbb{C}_{(n)}^{(13)} \cdot \mathbb{C}_{(13 - n)}^{(39)}$$

And the required probability is:

$$(2) \ \ \ \ \ \color{red} Prob(At \ Least \ 5 \ Hearts) \ = \ \frac { \sum_{n=5}^{13} \mathbb{C}_{(n)}^{(13)} \cdot \mathbb{C}_{(13 - n)}^{(39)} } { \mathbb{C}_{(13)}^{(52)} }$$

~~

Last edited: Mar 30, 2005
5. Mar 31, 2005

### spikemurphy

XANTHYM, Thanks so much for your help. When I saw your answer I checked my original message, the 13 in my jumbled formula was an error. I know that the cards have to add up to 52.

Thanks again, spikemurphy

p.s. is there an easy way to use latex, do I need a special program.

6. Mar 31, 2005