Discrete Mathematics

[thechampion116]

Question : Calculate the probability that a bridge hand contains exactly one ace? exactly two aces? only black cards?

The book tells us to use combinations

* I have one last question, i dont need an answer but someone to lead me in the right direction for calculation the chances of a 3-3-3-4 hand (suits.)

 

[pka]

Exactly one ace: (4)Combin(48,12)/Combin(52,13)
There are 4 aces and 48 other cards. Choose one ace and 12 others

 

[thechampion116]

thanks for the help, but could i get help for the others as well please

 

[pka]

You tell us what you think the other two should be!

 

[jolly]

With that response, it seems that this person just wants us to do his/her hw. Show us what you have tried.

 

[thechampion116]

For two aces, i tried pka's method but this time using 11 instead of 12 because i will have two aces. So my answer was (4) x Combin(48,11)/Combin(52,13), but i know i am still missing something

 

[pka]

Combin(4,2) x Combin(48,11)/Combin(52,13).
Choose 2 from 4.

 

[soroban]

Hello, thechampion116!

Calculate the probability that a bridge hand contains exactly one ace? exactly two aces? only black cards?

There are \(\displaystyle C(52,13)\,=\,\begin{pmatrix}52\\ 13\end{pmatrix}\,=\,\frac{52!}{13!\cdot39!}\) possible bridge hands.

To get 13 black cards, there are: \(\displaystyle C(26,13)\,=\,\begin{pmatrix}26\\13\end{pmatrix}\,=\,\frac{26!}{13!\cdot13!}\) ways.

Got it?

 

[thechampion116]

Thanks, i got the last one, but the answer listed in the book is 1.64 x 10^-4, and I am getting 1.64 x 10^-5, so i guess the book is wrong then.

 

[Lizzie1]

That has been know to happen every once in a while. But I would make sure.