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slides2 - Lecture Stat 302 Introduction to Probability...

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Lecture Stat 302 Introduction to Probability - Slides 2 AD Jan. 2010 AD () Jan. 2010 1 / 19
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Recapitulation Principle of counting : If experiment 1 has n 1 possible outcomes, experiment 2 has n 2 possible outcomes, .... , experiment r n r possible outcomes, then there is a total of n 1 ° n 2 ° ° ° ° ° n r possible outcomes of the r experiments. Permutations : For n objects, the number of permutations, i.e. ordered arrangements, of these n objects is given by n ! Permutations with objects alike : For n objects, of which n 1 are alike, n 2 are alike,..., n r are alike, there are n ! n 1 ! n 2 ! ° ° ° n r ! permutations. Combinations : The number of di/erent groups of r objects that can be formed from n objects is n ! ( n ± r ) ! if the order matters, n ! ( n ± r ) ! r ! if the order does not matter. AD () Jan. 2010 2 / 19
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Examples Example (Pb. 13) : Consider a group of 20 people. If everyone shakes hands with everyone else, how many handshakes take place? Answer : One handshake corresponds to one pair of people so we have to consider all possible pairs of people. There are ° 20 2 ± = 190 handshakes. Example (Pb. 15) : A dance class consists of 22 students, 10 W and 12 M. If 5 M and 5 W are to be chosen and then paired o/, how many results are possible? Answer : There are ° 10 5 ± ways to select 5 W and ° 12 5 ± ways to select 5 M. Then, given 5 W and 5 M, there are 5! possible permutations to combine them so the answer is ° 10 5 ± ° 12 5 ± 5 ! = 10 ! 5 ! 5 ! 12 ! 7 ! 5 ! 5 ! = 23 , 950 , 080 . AD () Jan. 2010 3 / 19
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Examples Example (Pb. 20) : A person has 8 friends, of whom 5 will be invited to a party. (a) How many choices are there if 2 of the friends are feuding and will not attend together? (b) How many choices if 2 of the friends will only attend together? Answer : (a) There are ° 8 5 ± = 56 possible groups of 5 friends if there were no constraint. Among those 56 groups, we have to exclude the ones where 2 of the friends are feuding. There are ° 2 2 ± ° 6 3 ± = 20 such groups so the answer is 56 ± 20 = 36 . (b) There are ° 2 2 ± ° 6 3 ± = 20 groups where the two friends attend together and ° 6 5 ± = 6 groups where none of them attend the party. So there are 20 + 6 = 26 possible groups. AD () Jan. 2010 4 / 19
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The Binomial Theorem Theorem We have ( x + y ) n = n k = 0 ° n k ± x k y n ± k . Example : Compute ( x + y ) 3 . Answer : We have ( x + y ) 3 = ° 3 0 ± x 0 y 3 + ° 3 1 ± xy 2 + ° 3 2 ± x 2 y + ° 3 3 ± x 3 y 0 = y 3 + 3 xy 2 + 3 x 2 y + x 3 . AD () Jan. 2010 5 / 19
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Proof of the binomial theorem by induction This is true at rank n = 1. Assume this is true at rank n ± 1 and let us prove it at rank n . We have ( x + y ) n = ( x + y ) ( x + y ) n ± 1 = ( x + y ) n ± 1 k = 0 ° n ± 1 k ± x k y n ± 1 ± k !
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