Math Is Fun Forum

Marakos

Guest

Probability Question HELP please

Question: Roll a fair die twice and find the probability of at least one 4 ???

Help me PLEASE

and thank you

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Probability Question HELP please

Hi Marakos;

If you mean a four on a single die then it's
11/36

Last edited by bobbym (2009-08-15 16:45:34)

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Probability Question HELP please

It's probably a bit easier to find the probability of getting no fours and then take that from 1.

The probability of throwing a non-four on a die is 5/6.
The probability of doing that twice is 5/6 * 5/6 = 25/36.
Hence the probability of not throwing no fours (ie. throwing at least one) is 1-25/36 = 11/36.

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Why did the vector cross the road?
It wanted to be normal.

Marakos

Guest

Re: Probability Question HELP please

Guys your my life savers..
Can you pelase explain to me one more problem..

Question: 
Thirteen cards are drawn at random without replacement from a standard deck of 52 cards. What is the probability that all are red ??

books staets C (26,13) / C(52/13)

I know that a deck has 26 Red cards but still I dont unserstand why they used C(26,13) ??

Why is 26 it should be 39

Identity

Member

Registered: 2007-04-18

Posts: 934

Re: Probability Question HELP please

C(26,13) is the number of ways of picking 13 objects from 26. There are 26 red cards in a deck, so C(26,13) is the number of ways of picking 13 cards from the 26 red cards.

Perhaps a better way to write it would be

At the top we are showing that we choose 13 cards from the 26 red cards, and 0 cards from the 26 black cards, i.e. all the cards we pick are red. The probability remains the same, since

.

C(52,13) is the number of ways of picking 13 cards from the entire deck.

Last edited by Identity (2009-08-15 16:13:18)

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Probability Question HELP please

Hi Marakos;

Here is another way to think about the problem, a little different than the books method.

For the first card you have 26 out of 52 cards for 26/52, for the second card you have 25 out of 51 for 25/51, each of these is multiplied.

Last edited by bobbym (2009-08-15 16:48:34)

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.