Randomly given assignment of some students to certain

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“randomly”: given assignment of some students to certain slots, any of the remaining students is equally likely to be assigned to any of the remaining slots. Question : What is the probability that each group includes a graduate student?
Multiplication Rule: Example 2 Denote the four graduate students by 1, 2, 3, 4 Define events ࠵? 1 = { students 1 and 2 are in different groups } , ࠵? 2 = { students 1, 2 and 3 are in different groups } , ࠵? 3 = { students 1, 2, 3 and 4 are in different groups } . We will use multiplication rule ࠵? ࠵? 3 = ࠵? ࠵? 1 ∩ ࠵? 2 ∩ ࠵? 3 = ࠵? ࠵? 1 ࠵? ࠵? 2 ࠵? 1 ࠵? ࠵? 3 ࠵? 1 ∩ ࠵? 2 ࠵? ࠵? 1 = 12/15 , ࠵? ࠵? 2 ࠵? 1 = 8/14 , ࠵? ࠵? 3 ࠵? 1 ∩ ࠵? 2 = 4/13 . So ࠵? ࠵? 3 = 12 15 8 14 4 13 ≈ 0.14 .
The Monty Hall Problem A prize is randomly put behind one of the three closed doors. You point to one door. A friend opens one of the remaining two doors, after making sure that the prize is not behind it. Question : Should you stick to your initial choice, or switch to the other unopened door?
The Monty Hall Problem If sticking to the initial choice: the initial choice determines whether you win or not, Thus the winning probability is 1/3. If switching to the other unopened door: Case 1: prize is behind the initial door, which happens with probability 1/3 . You don’t win. Case 2: prize is not behind the initial door, which happens with probability 2/3. You win for sure. So you should switch.
Content Sets. Probabilistic models. Conditional probability. Total Probability Theorem and Bayes’ Rule. Independence. Counting.
Total Probability Theorem Let ࠵? 1 , ࠵? 2 , . . . , ࠵? ࠵? be disjoint events that form a partition of the sample space. Assume ࠵?(࠵? ࠵? ) > 0 for all ࠵? . Then, for any event ࠵? , we have ࠵? ࠵? = ࠵? ࠵? 1 ∩ ࠵? + ⋯ + ࠵? ࠵? ࠵? ∩ ࠵? = ࠵? ࠵? 1 ࠵? ࠵? ࠵? 1 + ⋯ + ࠵? ࠵? ࠵? ࠵? ࠵? ࠵? ࠵? Indeed, ࠵? is the the disjoint union of ࠵? 1 ∩ ࠵? , , ࠵? ࠵? ∩ ࠵? . The second equality is given by ࠵? ࠵? ࠵? ∩ ࠵? = ࠵? ࠵? ࠵? ࠵? ࠵? ࠵? ࠵? .
Example: chess tournament Three types of players. Type 1: 50% Type 2: 25% Type 3: 25% You winning probability with these players: Against type 1: 0.3 . Against type 2: 0.4 . Against type 3: 0.5 . Now you play a game with a randomly chosen player. Question : What’s your winning probability?
Example: chess tournament ࠵? ࠵? : playing with an opponent of type ࠵? ࠵? ࠵? 1 = 0.5, ࠵? ࠵? 2 = 0.25, ࠵?(࠵? 3 ) = 0.25 . ࠵? : winning ࠵? ࠵? ࠵? 1 = 0.3, ࠵? ࠵? ࠵? 2 = 0.4, ࠵?(࠵?|࠵? 3 ) = 0.5 The probability of ࠵? : ࠵? ࠵? = ࠵? ࠵? 1 ࠵? ࠵? ࠵? 1 + ࠵? ࠵? 2 ࠵? ࠵? ࠵? 2 + ࠵?(࠵? 3 )࠵?(࠵?|࠵? 3 ) = 0.50 × 0.3 + 0.25 × 0.4 + 0.25 × 0.5 = 0.375
Example: Four-Sided Die Roll a fair 4-sided dice. Rule: Roll once more if result is 1 or 2, otherwise stop Question : What is the probability that the sum total of your rolls is at least 4?
Example: Four-Sided Die ࠵? ࠵? : the result of first roll is ࠵? ࠵? ࠵? ࠵? = 1/4 , ∀࠵? = 1,2,3,4. ࠵?: the sum total is at least 4. ࠵? ࠵? = σ ࠵?=1 4 ࠵? ࠵? ࠵? ࠵? ࠵? ࠵? ࠵? . Let’s calculate each ࠵? ࠵? ࠵? ࠵? . Given ࠵? 1 : the sum total will be ≥ 4 if the second roll results in 3 or 4, which happens with probability 1/2 . Thus ࠵?(࠵?|࠵? 1 ) = 1 2 , Similarly ࠵?(࠵?|࠵? 2 ) = 3 4 .

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