1. ## cards probability

A bag contains 8 cards numbered 1 through 8 .THe cards are mixed thoroughly and two cards are selected at random . Find the probability that both numbers are even .

I know this should be done this way ,

P(both numbers even)=(4C2)/(8C2)=3/14

My question is why couldn't it be done this way :

P(even)=1/2 so P(both number even)=(1/2)(1/2)=1/4 ??

2. Originally Posted by thereddevils
A bag contains 8 cards numbered 1 through 8 .THe cards are mixed thoroughly and two cards are selected at random . Find the probability that both numbers are even .

I know this should be done this way ,

P(both numbers even)=(4C2)/(8C2)=3/14

My question is why couldn't it be done this way :

P(even)=1/2 so P(both number even)=(1/2)(1/2)=1/4 ??
The probability that the second card is even given that the first is even is no longer 1/2.

The required probability is that the first card is one of the four even cards from a total of eight cards times that the second is one of the three remaining even cards of a total of seven cards = 3/14

CB

3. If you split the situation into

1. Take 2 cards at once

2. Take 1 card, pause and then take another

you will see the different ways you can calculate the same answer

What you did was to take a card, replace it, then take a card again

4. Originally Posted by Archie Meade
If you split the situation into

1. Take 2 cards at once

2. Take 1 card, pause and then take another

you will see the different ways you can calculate the same answer

What you did was to take a card, replace it, then take a card again
exactly , just realise that when i read captain's post . Thanks both of you !