Monday, September 30, 2013

Without Replacement




A player is dealt 2 cards from a standard deck of 52 cards.
What is the probability of getting a pair of aces?
 
Think about it....obviously this is an experiment "without replacement" because the player was given 2 cards.
To calculate the probability of a pair of aces you use the rules for compound events:
P(ace on first card) = 4/52
(remember,there are 4 aces in the deck)
P(ace on the second card) = 3/51!
(the first card drawn was an ace!)
So the probability of getting 2 aces is:
P(ace,ace) = (4/52)(3/51) = 12/2652 = 1/221!
 


In class I had trouble understanding the difference between Without Replacement and With Replacement. I will explain my learning experiences in another blog about understanding With Replacement. For now I would like to share what, and how I learned to deal with probability problems that deal with Without Replacement. 
So, I found this equation and I found it to be simple and helpful in understanding how exactly Without Replacement works. Using a standard deck of cards is the best way, in my opinion to understand the concept. The equation asked for the probability of getting a pair of aces. By drawing the first card, which is an ace and remember there are 4 aces in a deck, the probability is 4/52. Without Replacement comes into play by not putting that first drawn card back in the deck and simply drawing the next one from a deck of 53 cards.

Tuesday, September 24, 2013

Tree Diagrams

During the beginning of the course, we begin with Probabilities. When we were given the assignment to use a Tree Diagram, I was thrown off guard because I didn't know much about it. I found out that calculating probabilities isn't an easy task. Yet there are ways of solving a given problem, either with addition or multiplication.
After searching through a couple of sites, I found a particular site that gave me more information on what and how a Tree Diagram works. There is a Branch, Probability and an Outcome. Using these three, you calculate the probabilities by multiplying the first and second stages, and you would add all the outcomes together to reach the sum of 1