Card Problem probability urgent

[marvelous]

Hi
Here goes my question:

Suppose I pick 5 cards out of 52 , what is the probabilty of getting 2 cards same in those 5 cards ?

My soln : 13C(4,2)/C(52,5)

Can any one tell me if The solution is correct
Logic is : 13 suits in 52 cards .... picking 2 card out of 1 suit is C(4,2).... so total is 13C(4,2)
now picking 5 out of 52 is C(52,5)....

so solution is : 13C(4,2)/C(52,5)

can any confirm it?

Thanks

 

[Loren]

Suppose I pick 5 cards out of 52 , what is the probabilty of getting 2 cards same in those 5 cards ?

You need to rephrase your question. There are no two cards the same in a standard playing card deck. Two cards might be the same suit or two cards might be the same denomination. But they can't be both.

 

[soroban]

Hello, marvelous!

We need a more precise statement of the problem.

Suppose I pick 5 cards out of 52.
What is the probabilty of getting 2 cards same in those 5 cards ?

Your direction are not clear.
From your work, I assume you want two cards of the same value.
. . But your counting is off . . .


Do we want exactly one matching pair (and the other cards do not match at all)?
There are \(\displaystyle 13\) choices for the value of the pair.
Then there are \(\displaystyle C(4,2)\) ways to get the pair.

The third card can be any of the other 48 cards.
The fourth card can be any of the other 44 cards.
The fifth card can be any of the other 40 cards.
\(\displaystyle \text{Since their order does not matter, there are: }\:\frac{48\cdot44\cdot40}{3!} \:=\: 14,080\text{ ways.}\)

\(\displaystyle \text{Therefore, there are: }\:13 \times 6 \times 14,080 \;=\;1,098,240\,\text{ ways to get }exactly\text{ One Pair.}\)



Do we want at least a pair of matching values? .This would include Poker hands of:
. . One Pair, Two Pairs, Three of a Kind, Full House, and Four of a Kind.

\(\displaystyle \text{There are: }\:C(52,5) \:=\:2,598,960\text{ possible Poker hands.}\)


The opposite of "at least One Pair" is "NO Pairs."

The first card can be any of the 52 cards.
The second must be one of the other 48 cards.
The third must be one of the other 44 cards.
The fourth must be one of the other 40 cards.
The fifth must be one of the other 36 cards.

\(\displaystyle \text{Since their order does not matter, there are: }\:\frac{52\cdot48\cdot44\cdot40\cdot36}{5!} \:=\:1,317,888\text{ ways.}\)

\(\displaystyle \text{Therefore, there are: }\;2,598,960 - 1,317,888 \:=\:1,281,072\text{ ways to get }at\;least\text{ One Pair.}\)

 

[marvelous]

ok
Here goes exact problem :
I draw 5 cards from deck of cards .. what is the probability that 2 of the cards have same value?

thanks

 

[stapel]

Here goes exact problem :
I draw 5 cards from deck of cards .. what is the probability that 2 of the cards have same value?

Which part of the complete worked solution given to you does not fit the corrected exercise statement?

Please be specific, showing all of your work and reasoning (in particular, how your steps and values contradict the answer given to you). Thank you!

Eliz.