Conditional Probability

Quote:

Two cards are drawn at random from a pack of 8 cards numbered from 1 to 8. Find the conditional probability that:
(a) the sum of the two numbers is odd given that their product is even.
(b) the product of the two numbers is odd given that their sum is even.

First we note that
ExE -> E
ExO -> E
OxO -> O

E+E -> E
E+O -> O
O+O -> E

There are 4 even cards. We can choose any of 4, and then there are 3 left so we can choose any of 3. That gives 12 total EE pairs. But we don't care about the order so there are only 6 different pairs.

The same applies to the OO pairs. There are 6 different OO pairs.

The total number of different pairs is (8*7)/2 because we can choose any of 8, then any of 7, but again we don't care about the order, so there are 28 total different pairs.

Thus there must be 28 - 2*6 = 16 EO pairs

so in sum

6 different EE pairs
6 different OO pairs
16 different EO pairs

a) the product is even so we know we have either an EO or EE pair. Of those only the EO pairs give an odd sum.

so the conditional probability Pr[odd sum |even product]= (#EO pairs)/(#EO pairs + #EE pairs) = 16/(6+16) = 16/24 = 2/3

b) the sum is even here so we are looking at the EE and the OO pairs and we want to know how many of those the product is odd. The product being odd means it's only the OO pairs.

so the conditional probability Pr[odd product |even sum] = (#OO pairs)/(#EE pairs + #OO pairs) = 6/(6+6) = 1/2

bump
I really would like to look at this one .

bump
I really would like to look at this one .

tap_sheep:

Two cards are drawn at random from a pack of 8 cards numbered from 1 to 8. Find the conditional probability that:

(a) the sum of the two numbers is odd given that their product is even.
(b) the product of the two numbers is odd given that their sum is even.

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Thanks for your answer Rom. I am not as good at probablity as I should be and I got in a bit of a pickle when I tried this myself.
I am really just thinking allowed here. (as you said, order is not important anywhere here)

Total number of possible combinations = 8C2 = 28
Number of EE combinations = 4C2 = 6
Number of OO combinations = 4C2 = 6
Number of EO combinations = 4C1 * 4C1 = 4*4 = 16

(a) the sum of the two numbers is odd given that their product is even.
the product is even if the cards are EE or EO.
There are 6+16 = 22 of these combinations.
Of these combinations the sum will be odd if the cards are OE. There are 16 of these.
Therefore the probability that the sum of the two numbers is odd given that their product is even must be 16/22 = 8/11
(this is only different because you made a tiny adding error that tap_sheep could easily have picked up on him/herself. It is of no consequence.

b) Find the conditional probability that the product of the two numbers is odd given that their sum is even.
The sum is even if the numbers are EE or OO. There are 2+6=12 of these
The product is odd if the numbers are OO. There are 6 of these.
so the conditional probability that the product of the two numbers is odd given that their sum is even is 6/12 =1/2

Thank you for that lesson Rom.

I was actually thinking that maybe we could have a continuing thread dealing only with probability. I really need more practice. What do you think Rom?
Would anyone else like to comment or organise the thread?