If two cards are chosen at random from a standard deck of playing cards, what is the probability that at least one is an Ace or a King? Round your answer to two decimal places.

Question

If two cards are chosen at random from a standard deck of playing cards, what is the probability that at least one is an Ace or a King?   Round your answer to two decimal places.

amistre64 · Answer

2/13 perhaps?

amistre64 · Answer

ugh ... at least ..

anonymous · Answer

idk?

amistre64 · Answer

itd help if you had some idea :)

amistre64 · Answer

at least one ... means at most 2 .

amistre64 · Answer

maybe its better to see it as at most none ...

anonymous · Answer

yes that is the trick.  compute the probability that neither are.  then subtract from 1

amistre64 · Answer

52
-8
---
44

1 - 44/52 + 43/51 then?

anonymous · Answer

neither are is easy. first one isn't has probability 
$$\frac{44}{52}$$ then second one isn't given that the first one isn't has probability 
$$\frac{43}{51}$$ hope the reasoning is clear

anonymous · Answer

Well there's 52C2 ways to pick 2 cards, and 44C2 ways to not pick an A or K.  So your solution would be 1-(44C2/52C2) or something.

amistre64 · Answer

typoed meself lol

anonymous · Answer

so probability neither are is 
$$\frac{44}{52}\times \frac{43}{51}$$ and your answer is 
$$1-\frac{44}{52}\times \frac{43}{51}$$

anonymous · Answer

i am sure there is another way to do it, but it will come out to this answer.

anonymous · Answer

oh no never mind.  forget that

amistre64 · Answer

does the "or" mean addition?  or is that beside the point?

amistre64 · Answer

P(-A or -B)

anonymous · Answer

.29?

anonymous · Answer

that is what i got if you round

anonymous · Answer

@mooby your answer is the same as mine.  cancel the 2 in top and bottom and you get 
$$1-\frac{44\times 43}{52\times 51}$$