Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**genericname****Member**- Registered: 2012-05-16
- Posts: 52

Hi, I am a little confused about this problem:

*An urn contains four balls numbered 1,2,3 and 4. If two balls are drawn from the urn at random (that is, each pair has the same chance if being selected) and Z is the sum of the numbers on the two balls drawn, find:*

*a) the probability distribution of Z*

What I got for the answer was:

Urn= {1,2,3,4}

Z= {3,4,5,6,7}

p(3) = 1/4 //(1 way to get 3)

p(4) = 1/4 //(1 way to get 4)

p(5) = 1/2 //(2 ways to get 5)

p(6) = 1/4 //(1 way to get 6)

p(7) = 1/4 //(1 way to get 7)

Is this was the question was asking for? I wasn't quite sure. Though I think my answer is wrong because aren't they supposed to add up to 1?

Offline

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

Hi;

Though I think my answer is wrong because aren't they supposed to add up to 1?

I would like them very much to add up to one.

I get

2 ways to make a 3. ( 1 , 2 )and ( 2 , 1 ).

2 ways to make a 4. ( 1 , 3 )and ( 3 , 1 ).

4 ways to make a 5. ( 1 , 4 ),( 4, 1 ) and ( 2 , 3 )( 3 , 2 )

2 ways to make a 6. ( 2 , 4 )and ( 4 , 2 ).

2 ways to make a 7 ( 3 , 4 )and ( 4 , 3 ).

Can you finish now?

**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.**

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,168

hi genericname,

Have a look at my table below.

There are 12 possible outcomes so your probabilities should be /12

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**genericname****Member**- Registered: 2012-05-16
- Posts: 52

Oh, didn't know you can count them twice like that. Thank you for the replies.

Offline

**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,168

It's because you could draw a 1 followed by a 2 and also draw a 2 followed by a 1.

So you have to count both possibilities.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

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

Hi genericname;

You could have also counted them with order not counting as you did. As long as you are consistent.

**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.**

Offline

Pages: **1**