pick four cards randomly, what is the probability that you will get two distinct pairs?
@julay731 First of all welcome to openstudy : Given that total observations ( total selections ) ---> 4 In our favour ----> 2 probability -- > in favor / total --> 2/4 --> 1/2 May i know that r u doing Permutation ?
I should add that it is out of a deck of 52 cards. This is for a statistics class.
ok ! may be @Callisto help u best of luck :)
thank you
Since I am not sure with my answer..
Also the point that u added : deck of 52 cards will make no diff. in the ans.
okay no prob. thanks
\[([\left(\begin{matrix}4 \\ 1\end{matrix}\right)*\left(\begin{matrix}13 \\ 2\end{matrix}\right)+\left(\begin{matrix}3 \\ 1\end{matrix}\right)*\left(\begin{matrix}13 \\2\end{matrix}\right)]*4!)\div(\left(\begin{matrix}52 \\ 4\end{matrix}\right)*4!)\] That is select first colour and select two cards and then select the second colour and select the next two cards ==>(nominator) number of ways of selecting 2 pairs from a deck denominator==>(sample space) number of ways of selecting 4 cards from the deck
thank you that looks right. i need to review this.
it's not what I get
I would say....... probability is my enemy! 0% sure of my answer :S Assumption: the cards are (A, 2, 3, 4, 5 , 6, 7, 7, 8, 9, 10, J, Q, K) x 4 = 52 First, pick one card out of 52 cards => Probability = 52/52 = 1 Next pick another one to make a pair. => probability = 3/51 (only 3 cards with the same number or so are left after picking the first one) After that, randomly pick another card, condition: not same number as the first two => probability = (52-4) / 50 = 48/50 = 24/25 Last, pick a pick to form a pair. => probability = 3/49 Probability required = 1 x 3/51 x 24/25 x 3/49 = 72/20825 Epic fail.....
I don't get that either
What do you get?
\[\frac{{13 \choose 2}{4\choose 2}^2}{{52\choose 4}}\]
My Lord, do you mind explaining your answer???
I get everything except why did you square it?
there are 13 types of cards in order to form pairs with...choose 2 of them then from each pair chose 2 cards from the 4 thus \((\,_4C_2)^2\)
this problem is similar to finding the poker prob of getting 2 pairs with 5 cards... \[\frac{{13\choose 2}{4\choose 2}^2{11\choose 1}{4\choose 1}}{{52\choose 5}}\] only here we don't need that last card
Join our real-time social learning platform and learn together with your friends!