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**1a2b3c2212****Member**- Registered: 2009-04-04
- Posts: 419

Need some help with intepreting probability questions here.

1. 10 cards. Two numbered 2. Three numbered 3. Four numbered 4. 1 numbered 5. Draw two cards at the same time.

Find the probability that the 2 cards drawn cannot be a number 3 and a number 5.

-Are there like any hints are words to spot that can verify what kind of operation you must use?

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi 1a2b3c2212;

This is how I do it.

The number of ways to get a ( 3,5 ) = 3*1

The number of ways to get a ( 5,3 ) = 3*1

Number of ways to get (3,5) or (5,3) = 6

Total number of ways to draw 2 cards from the 10 is 10 * 9 = 90

6 / 90 = 1/15 is the probability of getting a 3 and a 5. The complement is the probability of not getting a 3 and a 5.

1 - 1/15 = 14/15

**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.**

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**qcao****Guest**

Alternative method:

P(first card = 3) = 3/10

P(second card = 5) = 1/9 because there are only 9 cards left to choose from.

P(3,5) = (3/10)(1/9) = 1/30

P(3,5 or 5,3) = 2(1/30) = 1/15

P(not 3 and 5) = 1 - 1/15 = 14/15

**John E. Franklin****Member**- Registered: 2005-08-29
- Posts: 3,588

Are you asking what if you don't get "3-3" or "3-5" combinations?

Try to make it clearer English?

**igloo** **myrtilles** **fourmis**

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