P b p a b proof clearly we can write the set b as 1 n

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P B P A B . Proof: Clearly we can write the set B as 1 ( ) n i i B P A B . Since the collection of i A form a partition, the collection of sets 1 ( ) n i i A B are disjoint sets. Rule 3 in our definition of probability says the probability of the union of disjoint sets equals the sum of the probabilities. The Law of Total Probability: To determine the probability of an event A, you sum up the probability of all ways that A can occur (loosely speaking). Example 2.32: An urn contains 8 white, 5 red and 3 blue chips. A person selects 3 chips without replacement. Determine the following probability: P(The third chip is red) One way to solve this problem is to determine all of the ways that the third chip is red. There are 4 ways that the third chip can be red. Thus, we have 3 1 2 3 1 2 3 1 2 3 1 2 3 ( ) ( ) ( ) ( ) ( ) 5 4 3 5 11 4 11 5 4 11 10 5 16 15 14 16 15 14 16 15 14 16 15 14 1050 5 .3125 3360 16 P R P R R R P R NR R P NR R R P NR NR R The second and third fraction in each part are determined by conditional probability. For the first term, a red on the first chip has probability 5/16. Once the first chip is red, we are down to 15 chips of which only 4 are red. So 2 1 ( | ) 4 /15 P R R . A simpler way to work THIS problem is to consider that you will pick the third chip first, set it aside and then pick the first two. Then the answer is more simply calculated as 5/16. This simple alternative solution should not be expected to be available on other problems. Example 2.33: Two cards are drawn from a standard deck of 52 cards. Determine the probability that the first card is a Heart and the second card is an Ace.
39 The simple method does not work on this problem. Here there are two things that need to take place and not just one thing as in Example 2.32 above (third chip red). Therefore, this problem needs the Law of Total Probability approach. The problem is written out in probability format as follows: 1 2 1 2 1 ( ) ( ) ( | ) P H A P H P A H . Determining 1 ( ) 13/ 52 P H is effortless. The second part seems to create some possible confusion. Clearly there are 51 cards left as the new whole. But, how many are Aces? Are there 3 Aces or 4 Aces. It depends if that first card, which was a heart, is an Ace. If it was, then there are 3 Aces left for the second cars, if the first card was not an Ace, then there are 4 Aces left. 1 2 1 2 1 2 ( ) ( ) ([ ] ) 1 3 12 4 51 1 52 51 52 51 52 51 52 P H A P A A P non- Ace A w w Example 2.34: An urn contains 7 white chips, 5 red chips and 3 blue chips. A chip is randomly selected from the urn, the color is noted, then chip is returned to the urn and then the number of chips of the noted color is doubled. Now a second chip is selected. Determine the probability that the first chip was white given that the second chip is blue. How should we start? 1 2 1 2 1 2 2 1 2 1 2 1 2 ( ) ( ) ( | ) ( ) ( ) ( ) ( ) P W B P W B P W B P B P W B P R B P B B 1 2 1 1 2 1 1 2 1 1 2 1 7 3 ( ) ( | ) 15 22 7 3 5 3 3 6 ( ) ( | ) ( ) ( | ) ( ) ( | ) 15 22 15 20 15 18 P W P B W P W P B W P R P B R P B P B B The first step above is the definition of conditional probability. Step 2 uses the law of total probability.

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