Question

In a standard deck, the cards are split evenly between four suits: hearts, diamonds, clubs, and spades. If you could take one card, put it back, and then take pick again, what is the probability that either the first card or the second card will be in the clubs suit?
40%?

Answer 1

Does having two clubs count? If so, then calculate the chances of you not getting any clubs at all.

(3/4)*(3/4)

And then 1-(that thing above)

Answer 2

I don't get it what?

Answer 3

Which part don't you get. Do you see how the chances of you NOT getting clubs at all is 3/4 times 3/4?

Answer 4

isnt the clubs be like 1/4?

Answer 5

yes. So the chances of you NOT getting clubs (you get diamonds spades or clover) is 3/4.

Answer 6

careful here

Answer 7

oh not ok

Answer 8

i read this question to say "one or the other but not both"

Answer 9

the compliment of getting no clubs is "first is a club, second is a club, or they are both clubs"

Answer 10

25% if its 3/4 odds?

Answer 11

if this means "one or the other or both" then lanT is right. but it it means "either the first or the second but not both" then we have to do something else

Answer 12

If it is not both then you use the first answer and subtract the chances that you DO get both, which is (1/4)*(1/4)

So, does having both count in your question?

Answer 13

probability you get it on the first pick AND not on the second is
\[\frac{1}{4}\times \frac{3}{4}=\frac{3}{16}\] and the probability you get it on the second AND not on the first is
\[\frac{3}{4}\times \frac{1}{4}=\frac{3}{16}\] so the probability you get it on the first AND not on the second OR you get it on the second AND not on the first is
\[\frac{3}{16}+\frac{3}{16}=\frac{9}{16}\]

Answer 14

@lanT you are of course right, and usually OR means one or the other or both, but in this case it looks like the inclusion of the word "either" indicates the exclusive or

Answer 15

still don't really get it is either added into it to confuse me?