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## #1 2010-02-09 23:37:50

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

### Probability

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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## #2 2010-02-10 03:21:29

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

### Re: Probability

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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## #3 2010-02-17 08:46:49

qcao
Guest

### Re: Probability

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

## #4 2010-02-17 10:26:41

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

### Re: Probability

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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