PERMUTATIONS and COMBINATIONS

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PERMUTATIONS and COMBINATIONS Powered By Docstoc
					                  .   Conditional Probabilities                .


 The probability of an event given that some other event has occurred.
i.e. a reduced sample space.
                                                         Sigma
                       
It is written as P B | A - the probability of B given A.
                                                         p161 Ex 9.01
From the tree diagram if A and B aren’t independent;

                                                    P(A  B)
P(A  B)  P(A)  PB | A           PB | A  
                                                      P(A)

E.g. Given that the toss of a die is even, what is the probability that it is
divisible by three?
        Method 1: Reduced sample space is {2,4,6}
                                                   1
               P(Divisible by 3 | toss is even) = 3
                                               1
                                    P(A  B)  6  1
        Method 2: Using PB | A   P(A)       1     3
                                                 2
E.g. The probability that a married man watches a certain TV show is 0.4
     The probability that a married woman watches the show is 0.5.
     The probability that a man watches the show given that his wife
does is 0.7.       Find the probability that

(i) A married couple both watch the show
(ii) A wife watches if her husband does
(iii) At least one person of a married couple watches the show

Answer      Let H = {husband watches show}
                W = {wife watches show}

(i) P(H  W)  P(H | W)  P(W)    0.5  0.7  0.35

                  P(H  W) 0.33
(ii) P(W | H)      P(H)
                          
                            0.4
                                 0.875


(iii) P(H  W)  P(H)  P(W)  P(H  W)
               = 0.4 + 0.5 – 0.35
               = 0.55
Sometimes the use of a table can make problems easier
E.g. In an examination 20% of students sitting failed Physics, 15%
failed Maths and 10% failed both. A student is selected at random

(a) If he failed Physics what is the probability that he failed Maths?
(b) If he failed Maths what is the probability that he failed Physics?
(c) What is the probability that he failed Maths or Physics?

Answer:
Let M = {passed Maths}         M′ = {failed Maths}         M    M′
    P = {passed Physics}       P′ = {failed Physics}   P 0.75 0.05 0.8
                                                       P′ 0.1 0.1 0.2
                 P( M  P) 0.1                          0.85 0.15 1
      P(M′|P′) =                  0.5
(a)
                    P(P )    0.2

                 P(M  P)    0.1
      P(P′|M′) =                   0.6667
(b)                P(M)      0.15
                                                            Sigma
(c) P(M  P)  P(M)  P(P)  P(M  P)                 p169 Ex 9.02
                = 0.15 + 0.2 – 0.1   = 0.25

				
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