Thread: Drawing from a deck of cards.

Suppose there is a deck consisting of 54 cards (standard deck with one black and one red joker). A single card is drawn from the deck of 54 cards but is not returned to the deck, and then a second card is drawn. Determine the probability of drawing:

1) the red joker or a red ace on either draw

I first thought it would be the probability of the first plus the probability of the second minus the intersection, but that was not the case.

The answer in the book is 1 - (50/53 * 51/54) = 52/477

Can anyone explain how they got their answer? Thanks!

Originally Posted by IanCarney

Suppose there is a deck consisting of 54 cards (standard deck with one black and one red joker). A single card is drawn from the deck of 54 cards but is not returned to the deck, and then a second card is drawn. Determine the probability of drawing:
1) the red joker or a red ace on either draw
The answer in the book is 1 - (50/53 * 51/54) = 52/477

There is one red joker and two red aces.
The probability of NEITHER a red joker nor a red ace is:
.
To get the complement, at least one, subtract that from 1.

Originally Posted by Plato

There is one red joker and two red aces.
The probability of NEITHER a red joker nor a red ace is:
.
To get the complement, at least one, subtract that from 1.

Is that the only way to do it?

Originally Posted by IanCarney

Is that the only way to do it?

Well of course it is not the only way.
But it is the cleanest and shortest way.
To be a good probability student, one must master this way of doing this kind of problem. You are finding the complement of the cases you do not want.

EDIT: This may interest you.


You should try to justify that fact.

Sorry to bump this yet again, but I think I confused myself upon looking at this again.

Last question: Why is the probability of not drawing a red joker or a red ace (51/54)(50/53)?

I know the (50/53) is from taking out the red joker and only having 3 aces to choose from, but why is it (51/54) instead of (53/54) if we're looking for the probability of not drawing a red joker?

Originally Posted by IanCarney

Why is the probability of not drawing a red joker or a red ace (51/54)(50/53)?

Remove the red joker and the two red aces.
That leaves 51 cards any two of which we can use.