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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.Quote:
Originally Posted by thereddevils
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
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 !Quote:
Originally Posted by Archie Meade