[TomProb]

Draw a number of cards with replacement from an ordinary deck of cards. What is the probability that it will require 'n' draws until all 4 suits are represented? As an example, say n=6.

[Alty]

Thank you for taking the time to copy your homework assignment on to our site.

Please read the following link regarding homework help.

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[Unknown008]

You have to put your part of the effort, TomProb.

What is the probability of drawing either a spade, a club, a diamond or a heart?
Then, what is the probability of drawing one of the three remaining, when you have drawn one of them.
After that, you do the same for the two suits remaining and finally for the last suit remaining.

With all these, you multiply each probability with the other, and you get your answer.

That gives you the probability for 4 draws, the minimum in that problem.

If you have 6 draws, that means you must have drawn a suit at least twice, and at most thrice before completing the required set.

[mathwiz3502]

There is probably a formula but it will be big or have crazy fractions

[Alty]

I have found that it is much better to keep your mouth shut and let people think you're a fool, than to open it and remove all doubt.

Then why didn't you keep your mouth shut?

Sounds like you have a few anger issues.

You may not be a student, but you posted a math question and 99.999% of the people that do are teens looking for answers to their homework.

We have strict rules on that here.

Why you felt the need to post what I quoted, I don't know. Maybe it's because you are a teen wanting us to do your homework and you're busted.

Either way, I don't take lip from anyone. I'm reporting this thread.

Good bye. :)

[Catsmine]

Couldn't guess your age, Tomprob, but your question is Algebra II homework straight from the book.

[galactus]

Hi Tom:

There is a 13/52=1/4 probability of drawing any suit from a 52-card deck.

With replacement, the first draw can be anything.

The second we want any other suit but the first one we drew: 39/52

The third draw would be any suit but the other two, and so on.

(39/52)(26/52)(13/52)=3/32

That is the probability of drawing the four suits in 4 draws.

Can you see how to adapt it to 'n' draws in general?

[Fr_Chuck]

I have deleted the smart mouth remarks of the OP, this is obvious homework, so perhaps a older person is gong back to school, but they also post in their anger very childish.

I am also closing this, since the OP is obvious too busy attacking to get any help.