<p>ok, so pretend that there are 26 cards in a box ok?</p>
<p>and these cards each have a pair. so there are 13 pairs of cards in a box, numbered from 1 to 13. (so there are two 1s, two 2s, two 3s...and so forth)</p>
<p>if two people randomly pick a card from the box, what is the probability that they will pick the same number card?</p>
<p>1/13 ?? (i might be wrong)</p>
<p>i'm pretty sure that's not right.</p>
<p>With replacement:
(26 C 1)*(2 C 1)/(26 C 2)</p>
<p>Without replacement:</p>
<p>(26 C 1) * (1 C 1)/(26 C 2)</p>
<p>god maybe thats 1/312 ...
ok thats crazy
what are the answer choices?</p>
<p>dude what does that even mean</p>
<p>can somebody give me an explanation in english (what the heck is meadow even saying?wth?)</p>
<p>and there are no answer choices, it is a grid in</p>
<p>oh than its deff not 1/312
ok i shut up lol</p>
<p>Okay, are you familiar with combinations? Well (n C r) is simply another way to say "how many ways can I pick r cards out n cards"? The basic way to approach most probability problems would be to list the (total possibilities given certain restraints )/( total possibilties in general)</p>
<p>so if there is replacement:</p>
<p>For the first person he/she has 13 total options of cards to pick from . Then in order for the second person to match the first card he/she was 2 possible cards to choose from. That being said there is a total number of possibilities is (26 C 2) since there 26 total cards and you need to choose 2. </p>
<p>Therefore I was actually slightly wrong in my original answer its actually:</p>
<p>(13 C 1)(2 C 1) / (26 C 2)</p>
<p>but if there is no replacement its:</p>
<p>(13 C 1)(1 C 1) / (26 C 2)</p>
<p>since there is no other card to pick but its pair</p>
<p>btw: you calculate n C r like this: n!/(n-r)!r!</p>
<p>all right i still didn't understand that, but can you tell me if i am right?</p>
<p>i thought that the probability of picking out a number card is 2/26
thus, the probability of picking out its pair is 1/25</p>
<p>so 2/26 x 1/25 = 1/325</p>
<p>^no that is still wrong. You need to find all the possibilities not such the probability of picking just 2 1's....</p>
<p>Just wondering, what don't you get?</p>
<p>i think its 2/26*1/25=1/325. becuz the first draw ur chances of taking lets say a "1" are 2/26 or 1/13. After taking that card away, there are only 25 cards left and the probability of taking another "1" would be 1/25. So then u multiply the probabilities together.</p>
<p>The answer is 1/25 (assuming that the first chooser does not put his card back in the deck before the next chooser picks).</p>
<p>Why? Because exactly two independent events must occur in succession: (1) first chooser picks a card (any card), and (2) second chooser picks the one remaining card in the deck with the identical number. Since first chooser can pick any card at all, the probability of that event is 1 (it is certain). Then only 1 of the remaining 25 cards has the same number, so the probability that the second chooser picks it is 1/25. The product of these two independent probabilities is 1/25.</p>
<p>^CorpsMan is right. Basically what I said in other words.</p>
<p>1/169?
chrrrrrr</p>
<p>@foo: what you found was the probability that a particular pair gets picked, say two #7 cards.
So, 1/25 is the probability of any pair, and the extra factor of 1/13 that you had gives you the probability for a particular pair.</p>
<p>In other words, the question tells u tht u can have any card in the first pick but in the second pick u will need the same card as in the first pick.
So ur prob. is (26/26) for the first pick.
And in the second pick u have 1/25
Therefore ur prob. is 1*(1/25)=1/25</p>