# Choosing numbers from a bag without replacement

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• Apr 23rd 2010, 12:13 PM
bhuang
Choosing numbers from a bag without replacement
A bag contains eleven discs numbered 1 to 11. Two discs are selected at random form the bag. Given that the sum of the selected numbers is even, use a tree diagram to find the probability that the number on each of the selected discs is odd.

Is there a way to do this problem withoutmaking an extensive tree diagram from 1 to 11?
• Apr 23rd 2010, 12:31 PM
Archie Meade
Quote:

Originally Posted by bhuang
A bag contains eleven discs numbered 1 to 11. Two discs are selected at random form the bag. Given that the sum of the selected numbers is even, use a tree diagram to find the probability that the number on each of the selected discs is odd.

Is there a way to do this problem withoutmaking an extensive tree diagram from 1 to 11?

Hi bhuang,

There are 6 odd numbers and 5 even numbers

Odd+Odd=Even and Even+Even=Even

$6c_2=\frac{6!}{(6-2)!2!}=15$

$5c_2=10$

there are 25 pairs of numbers that add to an even result.

The total number of pairs possible is $11c_2=55$

Therefore the probability of an even sum is $\frac{25}{55}$

To answer the question....

Given that the sum is Even, which means one of the 25 selections leading to an even result occurred,

in how many ways out of these 25 would the two numbers have been odd ?

Hence the probability is the fraction of that quantity out of the 25.
• Apr 23rd 2010, 01:02 PM
bhuang
I don't think that works because the answer is 3/5..
• Apr 23rd 2010, 01:16 PM
Archie Meade
Hi bhuang,

the probability of an even result isn't relevant to the question given.
The earlier post wasn't complete.
Left some calculations for you.

Additional work is added.

Two numbers add to an even result 25 times.
Two odd numbers add to an even result 15 times.

Given that the result was even,
the probability that the two numbers were odd is

$\frac{2\ odds\ sum\ to\ even\ result}{2\ numbers\ sum\ to\ even\ result}$