Probability of Getting at Least One Card from Each Suit

[Taylor_1989]

Hi all I am have a problem with this question I have used two method to solve and get two different ans, could someone please point where I am going wrong

Homework Statement

Five cards are dealt from a standard shuffled deck. What is the probability that:
There is at least one card from each of the four suits?[/B]

I wanted to see if I could attempt this question a different way to see if I could get the correct ans two ways. So what I did was say I pick 3 cards 2 from the same suit and one from another.

What I would like to know is my intuition correct because when I applied the same logic to the question above I got the same ans, I just want to make sure that I am not just getting the right ans from wrong workings

Now my ans for actual question which is in bold (13C2*13^3)/52C5

Homework Equations

The Attempt at a Solution

[/B]
So what I did was to say that if I pick 3 cards 2 of the same suit from a 52 card deck I could use combinations to calculate the probability:

(13C2*13C1)/52C3 this gve me an ans of 39/850

This is the ans if I just say picked hearts twice as there are four suits I need to multiply through by 4

Now I the attempted the question like this:

P(HHO)=13/52 *12/51 * 39/50= 39/850

I got a bit confused thinking to my self should I not work out P(HOH) and P(OHH) but then I re read the question half a dozen times think order dose not matter therefore I only need to calculate P(HHO) thus the ans are the same, I then applied the the same logic to the question above and got the same ans. Is this the right way about thinking about this question?

 

[mfb]

This is the ans if I just say picked hearts twice as there are four suits I need to multiply through by 4

Right. Where do you do that?

Now I the attempted the question like this:

P(HHO)=13/52 *12/51 * 39/50= 39/850

Those are just three cards where you care about the order, and I don't see the relevance to the problem statement.

 

[Ray Vickson]

Hi all I am have a problem with this question I have used two method to solve and get two different ans, could someone please point where I am going wrong

Homework Statement

Five cards are dealt from a standard shuffled deck. What is the probability that:
There is at least one card from each of the four suits?[/B]

I wanted to see if I could attempt this question a different way to see if I could get the correct ans two ways. So what I did was say I pick 3 cards 2 from the same suit and one from another.

What I would like to know is my intuition correct because when I applied the same logic to the question above I got the same ans, I just want to make sure that I am not just getting the right ans from wrong workings

Now my ans for actual question which is in bold (13C2*13^3)/52C5

Homework Equations

The Attempt at a Solution

[/B]
So what I did was to say that if I pick 3 cards 2 of the same suit from a 52 card deck I could use combinations to calculate the probability:

(13C2*13C1)/52C3 this gve me an ans of 39/850

This is the ans if I just say picked hearts twice as there are four suits I need to multiply through by 4

Now I the attempted the question like this:

P(HHO)=13/52 *12/51 * 39/50= 39/850

I got a bit confused thinking to my self should I not work out P(HOH) and P(OHH) but then I re read the question half a dozen times think order dose not matter therefore I only need to calculate P(HHO) thus the ans are the same, I then applied the the same logic to the question above and got the same ans. Is this the right way about thinking about this question?

The generalized hypergeometric distribution is useful here. If we have $N_1$ items of type 1, $N_2$ of type 2, ... $N_r$ of type $r$, with a total of $N = N_1 + N_2 + \cdots + N_r$ items, and we choose $n$ items at random without replacement, then the probability $p(k_1, k_2, \ldots, k_r)$ that we select $k_1$ items of type 1, $k_2$ of type 2, ... and $k_r$ of type $r$ (with $\sum k_i = n$) is
$$p(k_1, k_2, \ldots, k_r) = \frac{{N_1 \choose k_1} {N_2 \choose k_2} \cdots {N_r \choose k_r}}{{N \choose n}}$$
(Here, ${a \choose b} = aCb$.)
So, for example, the probability that you get 2 hearts and one each of spades, clubs and diamonds, is $p(2,1,1,1)$, computed with $N_1 = N_2 = N_3 = N_4 = 13, N = 52$ and $n = 5$.

 

[Taylor_1989]

@mfb
No I use the three card as an easier way for me to understand what was going on less number ect that all. This is the first time I have ever done probability besides basic stuff so I like make up an easier question on the same principle so I can understand what is going on.

 

[Ray Vickson]

@mfb
No I use the three card as an easier way for me to understand what was going on less number ect that all. This is the first time I have ever done probability besides basic stuff so I like make up an easier question on the same principle so I can understand what is going on.

This may be a poor strategy, if you are "making up" methods on your own that may (or may not) be wrong. You should check if your "simplified" way gives the same answer as the correct way.

As a word of warning: probability problems can be extremely tricky, and it is very easy to fall into a subtle trap. That is why I always recommend that beginners try to stay away from shortcuts, unless they are well-established shortcuts that have already been used many times by others (and appear in textbooks or papers, for example). I once read that probability is the subject in which more professional mathematicians have made errors that in any other field. That may be just a myth, but it serves as a useful reminder.

 

[Taylor_1989]

This may be a poor strategy, if you are "making up" methods on your own that may (or may not) be wrong. You should check if your "simplified" way gives the same answer as the correct way.

As a word of warning: probability problems can be extremely tricky, and it is very easy to fall into a subtle trap. That is why I always recommend that beginners try to stay away from shortcuts, unless they are well-established shortcuts that have already been used many times by others (and appear in textbooks or papers, for example). I once read that probability is the subject in which more professional mathematicians have made errors that in any other field. That may be just a myth, but it serves as a useful reminder.

Yes, I am just redoing my method now an realising that I am coming un stuck. I will stick using combinations as it make the most sense, I just wanted to see If I could apply this question to a generalised tree diagram and then workout from there. I just like to try different methods to is see if my ans correct. I am sure I have the correct ans by using the combination method I used above for the actual question?

 

[mfb]

(13C2*13^3)/52C5 is wrong.